Mixing
Which situations cause “Use of instruction is not an instruction” in LLVM IR?
0
I keep getting this error many times. Then, when I shift things around it works. My code is too complex to post here. I can't simplify and reproduce the problem.
The source code in LLVM that raises an exception is here: http://llvm.org/doxygen/Verifier_8cpp_source.html:
// Check that all uses of the instruction, if they are instructions
// themselves, actually have parent basic blocks. If the use is not an
// instruction, it is an error!
for (Use &U : I.uses()) {
if (Instruction *Used = dyn_cast<Instruction>(U.getUser()))
Assert(Used->getParent() != nullptr,
"Instruction referencing"
" instruction not embedded in a basic block!",
&I, Used);
else {
CheckFailed("Use of instruction is not an instruction!", U);
return;
}
}
But I still don't understand what it means
So, I wonder if anyone has a small example that cause the "Use of instruction is not an instruction" error and also explain why it happens.
Thank you!
llvm
asked Jan 1 at 9:33
TaninTanin
1,0441917
add a comment |
0
I keep getting this error many times. Then, when I shift things around it works. My code is too complex to post here. I can't simplify and reproduce the problem.
The source code in LLVM that raises an exception is here: http://llvm.org/doxygen/Verifier_8cpp_source.html:
// Check that all uses of the instruction, if they are instructions
// themselves, actually have parent basic blocks. If the use is not an
// instruction, it is an error!
for (Use &U : I.uses()) {
if (Instruction *Used = dyn_cast<Instruction>(U.getUser()))
Assert(Used->getParent() != nullptr,
"Instruction referencing"
" instruction not embedded in a basic block!",
&I, Used);
else {
CheckFailed("Use of instruction is not an instruction!", U);
return;
}
}
But I still don't understand what it means
So, I wonder if anyone has a small example that cause the "Use of instruction is not an instruction" error and also explain why it happens.
Thank you!
llvm
asked Jan 1 at 9:33
TaninTanin
1,0441917
1
Why exactly can't you make a smaller example from your current code? Your code is probably saved in a repository, which means you can delete some uninteresting part, try to compile it and if it still shows the error, even commit it to a local branch. Repeat until you have a small example.
– Roland Illig
Jan 1 at 9:55
I did that in multiple combinations. The problem disappeared and etc. It's definitely because I'm new to LLVM. So, I want take a different route of understanding this error by understanding a small example that causes it instead.
– Tanin
Jan 1 at 14:48
add a comment |
0
0
0
I keep getting this error many times. Then, when I shift things around it works. My code is too complex to post here. I can't simplify and reproduce the problem.
The source code in LLVM that raises an exception is here: http://llvm.org/doxygen/Verifier_8cpp_source.html:
// Check that all uses of the instruction, if they are instructions
// themselves, actually have parent basic blocks. If the use is not an
// instruction, it is an error!
for (Use &U : I.uses()) {
if (Instruction *Used = dyn_cast<Instruction>(U.getUser()))
Assert(Used->getParent() != nullptr,
"Instruction referencing"
" instruction not embedded in a basic block!",
&I, Used);
else {
CheckFailed("Use of instruction is not an instruction!", U);
return;
}
}
But I still don't understand what it means
So, I wonder if anyone has a small example that cause the "Use of instruction is not an instruction" error and also explain why it happens.
Thank you!
llvm
asked Jan 1 at 9:33
TaninTanin
1,0441917
I keep getting this error many times. Then, when I shift things around it works. My code is too complex to post here. I can't simplify and reproduce the problem.
The source code in LLVM that raises an exception is here: http://llvm.org/doxygen/Verifier_8cpp_source.html:
// Check that all uses of the instruction, if they are instructions
// themselves, actually have parent basic blocks. If the use is not an
// instruction, it is an error!
for (Use &U : I.uses()) {
if (Instruction *Used = dyn_cast<Instruction>(U.getUser()))
Assert(Used->getParent() != nullptr,
"Instruction referencing"
" instruction not embedded in a basic block!",
&I, Used);
else {
CheckFailed("Use of instruction is not an instruction!", U);
return;
}
}
But I still don't understand what it means
So, I wonder if anyone has a small example that cause the "Use of instruction is not an instruction" error and also explain why it happens.
Thank you!
llvm
llvm
asked Jan 1 at 9:33
TaninTanin
1,0441917
asked Jan 1 at 9:33
TaninTanin
1,0441917
asked Jan 1 at 9:33
TaninTanin
1,0441917
asked Jan 1 at 9:33
TaninTanin
1,0441917
asked Jan 1 at 9:33
TaninTanin
1,0441917
1,0441917
1
Why exactly can't you make a smaller example from your current code? Your code is probably saved in a repository, which means you can delete some uninteresting part, try to compile it and if it still shows the error, even commit it to a local branch. Repeat until you have a small example.
– Roland Illig
Jan 1 at 9:55
I did that in multiple combinations. The problem disappeared and etc. It's definitely because I'm new to LLVM. So, I want take a different route of understanding this error by understanding a small example that causes it instead.
– Tanin
Jan 1 at 14:48
add a comment |
1
Why exactly can't you make a smaller example from your current code? Your code is probably saved in a repository, which means you can delete some uninteresting part, try to compile it and if it still shows the error, even commit it to a local branch. Repeat until you have a small example.
– Roland Illig
Jan 1 at 9:55
I did that in multiple combinations. The problem disappeared and etc. It's definitely because I'm new to LLVM. So, I want take a different route of understanding this error by understanding a small example that causes it instead.
– Tanin
Jan 1 at 14:48
1
1
Why exactly can't you make a smaller example from your current code? Your code is probably saved in a repository, which means you can delete some uninteresting part, try to compile it and if it still shows the error, even commit it to a local branch. Repeat until you have a small example.
– Roland Illig
Jan 1 at 9:55
Why exactly can't you make a smaller example from your current code? Your code is probably saved in a repository, which means you can delete some uninteresting part, try to compile it and if it still shows the error, even commit it to a local branch. Repeat until you have a small example.
– Roland Illig
Jan 1 at 9:55
I did that in multiple combinations. The problem disappeared and etc. It's definitely because I'm new to LLVM. So, I want take a different route of understanding this error by understanding a small example that causes it instead.
– Tanin
Jan 1 at 14:48
I did that in multiple combinations. The problem disappeared and etc. It's definitely because I'm new to LLVM. So, I want take a different route of understanding this error by understanding a small example that causes it instead.
– Tanin
Jan 1 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
2
You are probably generating bogus IR somewhere in your code. The assertion you mention might be triggered by, for example, constructing a constant expression that uses result of other instruction as operand:
%0 = <some instruction producing result>
store i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0), ...
Here, the getelementptr inbounds ... part is a constant expression, which can't contain anything except other constants. Hence, we can't use %0 as indices here.
Instead, we need to use getelementptr instruction:
%0 = <some instruction producing result>
%1 = i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0)
store i8* %1, ...
As for your case, you can just call I.dump() from your code or inside debugger to figure out what instruction causes failed assertion exactly.
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
For my specific problem, I'm stuck with this:Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.
– Tanin
Jan 1 at 23:15
I've figured it out. I later used%size_with_terminatorin another constant expression. The error was misleading; It was the use of%size_with_terminator, not the declaration of%size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.
– Tanin
Jan 1 at 23:45
1
I'm using debugger for this. All ofInstruction,BasicBlock,FunctionandModuleclasses havedump()method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.
– arrowd
Jan 2 at 8:59
|
show 1 more comment
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1 Answer
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2
You are probably generating bogus IR somewhere in your code. The assertion you mention might be triggered by, for example, constructing a constant expression that uses result of other instruction as operand:
%0 = <some instruction producing result>
store i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0), ...
Here, the getelementptr inbounds ... part is a constant expression, which can't contain anything except other constants. Hence, we can't use %0 as indices here.
Instead, we need to use getelementptr instruction:
%0 = <some instruction producing result>
%1 = i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0)
store i8* %1, ...
As for your case, you can just call I.dump() from your code or inside debugger to figure out what instruction causes failed assertion exactly.
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
For my specific problem, I'm stuck with this:Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.
– Tanin
Jan 1 at 23:15
I've figured it out. I later used%size_with_terminatorin another constant expression. The error was misleading; It was the use of%size_with_terminator, not the declaration of%size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.
– Tanin
Jan 1 at 23:45
1
I'm using debugger for this. All ofInstruction,BasicBlock,FunctionandModuleclasses havedump()method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.
– arrowd
Jan 2 at 8:59
|
show 1 more comment
2
You are probably generating bogus IR somewhere in your code. The assertion you mention might be triggered by, for example, constructing a constant expression that uses result of other instruction as operand:
%0 = <some instruction producing result>
store i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0), ...
Here, the getelementptr inbounds ... part is a constant expression, which can't contain anything except other constants. Hence, we can't use %0 as indices here.
Instead, we need to use getelementptr instruction:
%0 = <some instruction producing result>
%1 = i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0)
store i8* %1, ...
As for your case, you can just call I.dump() from your code or inside debugger to figure out what instruction causes failed assertion exactly.
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
For my specific problem, I'm stuck with this:Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.
– Tanin
Jan 1 at 23:15
I've figured it out. I later used%size_with_terminatorin another constant expression. The error was misleading; It was the use of%size_with_terminator, not the declaration of%size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.
– Tanin
Jan 1 at 23:45
1
I'm using debugger for this. All ofInstruction,BasicBlock,FunctionandModuleclasses havedump()method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.
– arrowd
Jan 2 at 8:59
|
show 1 more comment
2
2
2
You are probably generating bogus IR somewhere in your code. The assertion you mention might be triggered by, for example, constructing a constant expression that uses result of other instruction as operand:
%0 = <some instruction producing result>
store i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0), ...
Here, the getelementptr inbounds ... part is a constant expression, which can't contain anything except other constants. Hence, we can't use %0 as indices here.
Instead, we need to use getelementptr instruction:
%0 = <some instruction producing result>
%1 = i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0)
store i8* %1, ...
As for your case, you can just call I.dump() from your code or inside debugger to figure out what instruction causes failed assertion exactly.
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
You are probably generating bogus IR somewhere in your code. The assertion you mention might be triggered by, for example, constructing a constant expression that uses result of other instruction as operand:
%0 = <some instruction producing result>
store i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0), ...
Here, the getelementptr inbounds ... part is a constant expression, which can't contain anything except other constants. Hence, we can't use %0 as indices here.
Instead, we need to use getelementptr instruction:
%0 = <some instruction producing result>
%1 = i8* getelementptr inbounds ([123 x i8], [123 x i8]* @some_string, i32 %0, i32 0)
store i8* %1, ...
As for your case, you can just call I.dump() from your code or inside debugger to figure out what instruction causes failed assertion exactly.
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
answered Jan 1 at 19:33
arrowdarrowd
22.7k45383
22.7k45383
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
For my specific problem, I'm stuck with this:Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.
– Tanin
Jan 1 at 23:15
I've figured it out. I later used%size_with_terminatorin another constant expression. The error was misleading; It was the use of%size_with_terminator, not the declaration of%size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.
– Tanin
Jan 1 at 23:45
1
I'm using debugger for this. All ofInstruction,BasicBlock,FunctionandModuleclasses havedump()method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.
– arrowd
Jan 2 at 8:59
|
show 1 more comment
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
For my specific problem, I'm stuck with this:Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.
– Tanin
Jan 1 at 23:15
I've figured it out. I later used%size_with_terminatorin another constant expression. The error was misleading; It was the use of%size_with_terminator, not the declaration of%size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.
– Tanin
Jan 1 at 23:45
1
I'm using debugger for this. All ofInstruction,BasicBlock,FunctionandModuleclasses havedump()method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.
– arrowd
Jan 2 at 8:59
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
Let me get this right. We merely need to store the result in an intermediate variable to make it a non-constant, right? Coming from a normal programming language, this is a huge gotcha. I didn't even realise the above 2 code were not equivalent. Could you point to more explanation around this concept? What is this concept called? I'd like to read more.
– Tanin
Jan 1 at 22:44
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
More question: I notice that it only fails when verifying the function. What happens if we just go ahead and run the code anyway? How would it be wrong? Thank you!
– Tanin
Jan 1 at 22:51
For my specific problem, I'm stuck with this:
Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.– Tanin
Jan 1 at 23:15
For my specific problem, I'm stuck with this:
Use of instruction is not an instruction! %size_with_terminator = add i32 %cstring_size, 1. I'm not sure how to solve it. The line is already on its own.– Tanin
Jan 1 at 23:15
I've figured it out. I later used
%size_with_terminator in another constant expression. The error was misleading; It was the use of %size_with_terminator, not the declaration of %size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.– Tanin
Jan 1 at 23:45
I've figured it out. I later used
%size_with_terminator in another constant expression. The error was misleading; It was the use of %size_with_terminator, not the declaration of %size_with_terminator. Do you have any trick to figure this out quickly? It failed when verifying a function, so I couldn't even run it. Thank you for the pointer. It really helps me understand LLVM better.– Tanin
Jan 1 at 23:45
1
1
I'm using debugger for this. All of
Instruction, BasicBlock, Function and Module classes have dump() method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.– arrowd
Jan 2 at 8:59
I'm using debugger for this. All of
Instruction, BasicBlock, Function and Module classes have dump() method that prints corresponding IR. Whenever assertion fails, the debuggers breaks and allows you to inspect IR on which Verifier choked.– arrowd
Jan 2 at 8:59
|
show 1 more comment
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1
Why exactly can't you make a smaller example from your current code? Your code is probably saved in a repository, which means you can delete some uninteresting part, try to compile it and if it still shows the error, even commit it to a local branch. Repeat until you have a small example.
– Roland Illig
Jan 1 at 9:55
I did that in multiple combinations. The problem disappeared and etc. It's definitely because I'm new to LLVM. So, I want take a different route of understanding this error by understanding a small example that causes it instead.
– Tanin
Jan 1 at 14:48