Mixing
When I am going to multiply uint32 variable with sint32 variable why I am getting wrong ans?
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sint32 MulDiv (uint32 x_value, sint32 y_value, uint32 z_value) {
sint64 v_Result_value;
sint64 v_Temp_Val;
v_Temp_Val = x_value * y_value;
if (0 == z_value) {
if(v_Temp_Val >= 0) {
v_Result_value = MAX_VAL;
} else {
v_Result_value = MIN_VAL;
}
}
return v_Result_value;
}
If I pass:
- x_value as 4294967295
- y_value as -12
v_Temp_Val become 12, which is not expected.
c embedded
edited Jan 3 at 10:05
CristiFati
15k72538
asked Jan 3 at 9:53
TejasTejas
126
add a comment |
sint32 MulDiv (uint32 x_value, sint32 y_value, uint32 z_value) {
sint64 v_Result_value;
sint64 v_Temp_Val;
v_Temp_Val = x_value * y_value;
if (0 == z_value) {
if(v_Temp_Val >= 0) {
v_Result_value = MAX_VAL;
} else {
v_Result_value = MIN_VAL;
}
}
return v_Result_value;
}
If I pass:
- x_value as 4294967295
- y_value as -12
v_Temp_Val become 12, which is not expected.
c embedded
edited Jan 3 at 10:05
CristiFati
15k72538
asked Jan 3 at 9:53
TejasTejas
126
What ifv_Temp_Val = (x_value * (sint64)y_value);?
– CristiFati
Jan 3 at 10:00
........and what did you pass as z? Please indent your code clearly to avoid misinterpretation of control flow.
– Martin James
Jan 3 at 10:02
add a comment |
sint32 MulDiv (uint32 x_value, sint32 y_value, uint32 z_value) {
sint64 v_Result_value;
sint64 v_Temp_Val;
v_Temp_Val = x_value * y_value;
if (0 == z_value) {
if(v_Temp_Val >= 0) {
v_Result_value = MAX_VAL;
} else {
v_Result_value = MIN_VAL;
}
}
return v_Result_value;
}
If I pass:
- x_value as 4294967295
- y_value as -12
v_Temp_Val become 12, which is not expected.
c embedded
edited Jan 3 at 10:05
CristiFati
15k72538
asked Jan 3 at 9:53
TejasTejas
126
sint32 MulDiv (uint32 x_value, sint32 y_value, uint32 z_value) {
sint64 v_Result_value;
sint64 v_Temp_Val;
v_Temp_Val = x_value * y_value;
if (0 == z_value) {
if(v_Temp_Val >= 0) {
v_Result_value = MAX_VAL;
} else {
v_Result_value = MIN_VAL;
}
}
return v_Result_value;
}
If I pass:
- x_value as 4294967295
- y_value as -12
v_Temp_Val become 12, which is not expected.
c embedded
c embedded
edited Jan 3 at 10:05
CristiFati
15k72538
asked Jan 3 at 9:53
TejasTejas
126
edited Jan 3 at 10:05
CristiFati
15k72538
asked Jan 3 at 9:53
TejasTejas
126
edited Jan 3 at 10:05
CristiFati
15k72538
edited Jan 3 at 10:05
CristiFati
15k72538
edited Jan 3 at 10:05
CristiFati
15k72538
15k72538
asked Jan 3 at 9:53
TejasTejas
126
asked Jan 3 at 9:53
TejasTejas
126
asked Jan 3 at 9:53
TejasTejas
126
126
What ifv_Temp_Val = (x_value * (sint64)y_value);?
– CristiFati
Jan 3 at 10:00
........and what did you pass as z? Please indent your code clearly to avoid misinterpretation of control flow.
– Martin James
Jan 3 at 10:02
add a comment |
What ifv_Temp_Val = (x_value * (sint64)y_value);?
– CristiFati
Jan 3 at 10:00
........and what did you pass as z? Please indent your code clearly to avoid misinterpretation of control flow.
– Martin James
Jan 3 at 10:02
What if
v_Temp_Val = (x_value * (sint64)y_value);?– CristiFati
Jan 3 at 10:00
What if
v_Temp_Val = (x_value * (sint64)y_value);?– CristiFati
Jan 3 at 10:00
........and what did you pass as z? Please indent your code clearly to avoid misinterpretation of control flow.
– Martin James
Jan 3 at 10:02
........and what did you pass as z? Please indent your code clearly to avoid misinterpretation of control flow.
– Martin James
Jan 3 at 10:02
add a comment |
1 Answer
1
active
oldest
votes
With the expression x_value * y_value, the values of x_value and y_value goes through Usual Arithmetic Conversion, which
Otherwise, the signedness is different: If the operand with the unsigned type has conversion rank greater or equal than the rank of the type of the signed operand, then the operand with the signed type is implicitly converted to the unsigned type
That is, the value -12 is "converted" to an unsigned value, and due to how two's complement (the most common way to handle negative numbers on computers) works the value -12 is converted to a very large value (4294967284 more precisely).
Multiplying 4294967295 by 4294967284 result in arithmetic overflow (because it's a 32-bit multiplication with a 32-bit result), but it's well-defined for unsigned integers. However the result will not be what you expect.
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
With the expression x_value * y_value, the values of x_value and y_value goes through Usual Arithmetic Conversion, which
Otherwise, the signedness is different: If the operand with the unsigned type has conversion rank greater or equal than the rank of the type of the signed operand, then the operand with the signed type is implicitly converted to the unsigned type
That is, the value -12 is "converted" to an unsigned value, and due to how two's complement (the most common way to handle negative numbers on computers) works the value -12 is converted to a very large value (4294967284 more precisely).
Multiplying 4294967295 by 4294967284 result in arithmetic overflow (because it's a 32-bit multiplication with a 32-bit result), but it's well-defined for unsigned integers. However the result will not be what you expect.
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
add a comment |
With the expression x_value * y_value, the values of x_value and y_value goes through Usual Arithmetic Conversion, which
Otherwise, the signedness is different: If the operand with the unsigned type has conversion rank greater or equal than the rank of the type of the signed operand, then the operand with the signed type is implicitly converted to the unsigned type
That is, the value -12 is "converted" to an unsigned value, and due to how two's complement (the most common way to handle negative numbers on computers) works the value -12 is converted to a very large value (4294967284 more precisely).
Multiplying 4294967295 by 4294967284 result in arithmetic overflow (because it's a 32-bit multiplication with a 32-bit result), but it's well-defined for unsigned integers. However the result will not be what you expect.
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
add a comment |
With the expression x_value * y_value, the values of x_value and y_value goes through Usual Arithmetic Conversion, which
Otherwise, the signedness is different: If the operand with the unsigned type has conversion rank greater or equal than the rank of the type of the signed operand, then the operand with the signed type is implicitly converted to the unsigned type
That is, the value -12 is "converted" to an unsigned value, and due to how two's complement (the most common way to handle negative numbers on computers) works the value -12 is converted to a very large value (4294967284 more precisely).
Multiplying 4294967295 by 4294967284 result in arithmetic overflow (because it's a 32-bit multiplication with a 32-bit result), but it's well-defined for unsigned integers. However the result will not be what you expect.
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
With the expression x_value * y_value, the values of x_value and y_value goes through Usual Arithmetic Conversion, which
Otherwise, the signedness is different: If the operand with the unsigned type has conversion rank greater or equal than the rank of the type of the signed operand, then the operand with the signed type is implicitly converted to the unsigned type
That is, the value -12 is "converted" to an unsigned value, and due to how two's complement (the most common way to handle negative numbers on computers) works the value -12 is converted to a very large value (4294967284 more precisely).
Multiplying 4294967295 by 4294967284 result in arithmetic overflow (because it's a 32-bit multiplication with a 32-bit result), but it's well-defined for unsigned integers. However the result will not be what you expect.
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
answered Jan 3 at 10:06
Some programmer dudeSome programmer dude
305k25265427
305k25265427
add a comment |
add a comment |
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What if
v_Temp_Val = (x_value * (sint64)y_value);?– CristiFati
Jan 3 at 10:00
........and what did you pass as z? Please indent your code clearly to avoid misinterpretation of control flow.
– Martin James
Jan 3 at 10:02