Mixing
Getting substrings in C [closed]
I currently have this string called'json' in C "{"method":"emptyFood","params":"08:00 09:00 10:00"}"
and trying to extract the times into 3 different variables.
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
for(int j = 0; j<5; j++)
{
ts2[j] = json[j+38];
}
for (int k = 0; k<5; k++)
{
ts3[k] = json[k+44];
}
printf("time1 = %sn", ts1);
printf("time2 = %sn", ts2);
printf("time3 = %sn", ts3);
If I print them all at the same time, I get strange outputs like this
time1 = 08:009:010:00
time2 = 09:010:00
time3 = 10:00
However if I print them one by one, it prints out the correct time for each print statement.
What am I doing wrong?
c string
asked Nov 21 '18 at 21:32
HerofireHerofire
416
closed as off-topic by Jonathan Leffler, Weather Vane, chux, Unheilig, EdChum Nov 22 '18 at 9:32
This question appears to be off-topic. The users who voted to close gave these specific reasons:
- "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Jonathan Leffler, EdChum
- "Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See: How to create a Minimal, Complete, and Verifiable example." – Weather Vane, chux
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 2 more comments
I currently have this string called'json' in C "{"method":"emptyFood","params":"08:00 09:00 10:00"}"
and trying to extract the times into 3 different variables.
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
for(int j = 0; j<5; j++)
{
ts2[j] = json[j+38];
}
for (int k = 0; k<5; k++)
{
ts3[k] = json[k+44];
}
printf("time1 = %sn", ts1);
printf("time2 = %sn", ts2);
printf("time3 = %sn", ts3);
If I print them all at the same time, I get strange outputs like this
time1 = 08:009:010:00
time2 = 09:010:00
time3 = 10:00
However if I print them one by one, it prints out the correct time for each print statement.
What am I doing wrong?
c string
asked Nov 21 '18 at 21:32
HerofireHerofire
416
closed as off-topic by Jonathan Leffler, Weather Vane, chux, Unheilig, EdChum Nov 22 '18 at 9:32
This question appears to be off-topic. The users who voted to close gave these specific reasons:
- "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Jonathan Leffler, EdChum
- "Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See: How to create a Minimal, Complete, and Verifiable example." – Weather Vane, chux
If this question can be reworded to fit the rules in the help center, please edit the question.
6
You aren't zero-terminating your strings.
– Lee Daniel Crocker
Nov 21 '18 at 21:36
Can you show us "printing them all at the same time" vs "printing them 1 by 1"?
– CoffeeTableEspresso
Nov 21 '18 at 21:36
Can you show the target string definitions?
– Weather Vane
Nov 21 '18 at 21:36
3
Please include a main() function and provide a Compilable, Minimal, Complete, and Verifiable Example. It makes it a lot easier to reproduce what you are seeing.
– John Murray
Nov 21 '18 at 21:43
2
"What am I doing wrong?" --> You are not posting the definitions ofts1, ts2, ts3
– chux
Nov 21 '18 at 21:55
|
show 2 more comments
I currently have this string called'json' in C "{"method":"emptyFood","params":"08:00 09:00 10:00"}"
and trying to extract the times into 3 different variables.
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
for(int j = 0; j<5; j++)
{
ts2[j] = json[j+38];
}
for (int k = 0; k<5; k++)
{
ts3[k] = json[k+44];
}
printf("time1 = %sn", ts1);
printf("time2 = %sn", ts2);
printf("time3 = %sn", ts3);
If I print them all at the same time, I get strange outputs like this
time1 = 08:009:010:00
time2 = 09:010:00
time3 = 10:00
However if I print them one by one, it prints out the correct time for each print statement.
What am I doing wrong?
c string
asked Nov 21 '18 at 21:32
HerofireHerofire
416
I currently have this string called'json' in C "{"method":"emptyFood","params":"08:00 09:00 10:00"}"
and trying to extract the times into 3 different variables.
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
for(int j = 0; j<5; j++)
{
ts2[j] = json[j+38];
}
for (int k = 0; k<5; k++)
{
ts3[k] = json[k+44];
}
printf("time1 = %sn", ts1);
printf("time2 = %sn", ts2);
printf("time3 = %sn", ts3);
If I print them all at the same time, I get strange outputs like this
time1 = 08:009:010:00
time2 = 09:010:00
time3 = 10:00
However if I print them one by one, it prints out the correct time for each print statement.
What am I doing wrong?
c string
c string
asked Nov 21 '18 at 21:32
HerofireHerofire
416
asked Nov 21 '18 at 21:32
HerofireHerofire
416
asked Nov 21 '18 at 21:32
HerofireHerofire
416
asked Nov 21 '18 at 21:32
HerofireHerofire
416
asked Nov 21 '18 at 21:32
HerofireHerofire
416
416
closed as off-topic by Jonathan Leffler, Weather Vane, chux, Unheilig, EdChum Nov 22 '18 at 9:32
This question appears to be off-topic. The users who voted to close gave these specific reasons:
- "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Jonathan Leffler, EdChum
- "Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See: How to create a Minimal, Complete, and Verifiable example." – Weather Vane, chux
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jonathan Leffler, Weather Vane, chux, Unheilig, EdChum Nov 22 '18 at 9:32
This question appears to be off-topic. The users who voted to close gave these specific reasons:
- "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Jonathan Leffler, EdChum
- "Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See: How to create a Minimal, Complete, and Verifiable example." – Weather Vane, chux
If this question can be reworded to fit the rules in the help center, please edit the question.
6
You aren't zero-terminating your strings.
– Lee Daniel Crocker
Nov 21 '18 at 21:36
Can you show us "printing them all at the same time" vs "printing them 1 by 1"?
– CoffeeTableEspresso
Nov 21 '18 at 21:36
Can you show the target string definitions?
– Weather Vane
Nov 21 '18 at 21:36
3
Please include a main() function and provide a Compilable, Minimal, Complete, and Verifiable Example. It makes it a lot easier to reproduce what you are seeing.
– John Murray
Nov 21 '18 at 21:43
2
"What am I doing wrong?" --> You are not posting the definitions ofts1, ts2, ts3
– chux
Nov 21 '18 at 21:55
|
show 2 more comments
6
You aren't zero-terminating your strings.
– Lee Daniel Crocker
Nov 21 '18 at 21:36
Can you show us "printing them all at the same time" vs "printing them 1 by 1"?
– CoffeeTableEspresso
Nov 21 '18 at 21:36
Can you show the target string definitions?
– Weather Vane
Nov 21 '18 at 21:36
3
Please include a main() function and provide a Compilable, Minimal, Complete, and Verifiable Example. It makes it a lot easier to reproduce what you are seeing.
– John Murray
Nov 21 '18 at 21:43
2
"What am I doing wrong?" --> You are not posting the definitions ofts1, ts2, ts3
– chux
Nov 21 '18 at 21:55
6
6
You aren't zero-terminating your strings.
– Lee Daniel Crocker
Nov 21 '18 at 21:36
You aren't zero-terminating your strings.
– Lee Daniel Crocker
Nov 21 '18 at 21:36
Can you show us "printing them all at the same time" vs "printing them 1 by 1"?
– CoffeeTableEspresso
Nov 21 '18 at 21:36
Can you show us "printing them all at the same time" vs "printing them 1 by 1"?
– CoffeeTableEspresso
Nov 21 '18 at 21:36
Can you show the target string definitions?
– Weather Vane
Nov 21 '18 at 21:36
Can you show the target string definitions?
– Weather Vane
Nov 21 '18 at 21:36
3
3
Please include a main() function and provide a Compilable, Minimal, Complete, and Verifiable Example. It makes it a lot easier to reproduce what you are seeing.
– John Murray
Nov 21 '18 at 21:43
Please include a main() function and provide a Compilable, Minimal, Complete, and Verifiable Example. It makes it a lot easier to reproduce what you are seeing.
– John Murray
Nov 21 '18 at 21:43
2
2
"What am I doing wrong?" --> You are not posting the definitions of
ts1, ts2, ts3– chux
Nov 21 '18 at 21:55
"What am I doing wrong?" --> You are not posting the definitions of
ts1, ts2, ts3– chux
Nov 21 '18 at 21:55
|
show 2 more comments
1 Answer
1
active
oldest
votes
Strings in C are null-terminated. This means that a string ends with a null byte. (A byte in a string has the type char.) When you copy the bytes of a substring to a new array of bytes, you're copying the “useful” content of the string, but to get a complete string, you also need to write the null terminator. For example:
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
ts1[5] = 0;
(Don't confuse the null byte 0, which can also be written '', with the digit '0'.)
A more idiomatic way of writing this would be
memcpy(ts1, json + 32, 5);
ts1[5] = 0;
A better-designed function to copy a substring is strlcpy but unfortunately it is not part of standard C so many platforms don't have it. Note that strncpy doesn't do the job well because it wouldn't write a null byte in your case.
Note that you need 6 bytes to store a 5-byte string, i.e. you must declare char ts1[6] or char *ts1 = malloc(6) (or with a larger number).
The reason it seemed to work when you printed the strings one by one is that there must have been a null byte immediately after the string in memory. This is just a coincidence and might not happen if the function was called from different place, if the code before it had worked on different data, if compiled with a different compiler, etc. Given the output you got with the three variables, what must have happened is that the compiler happened to put ts1, ts2 and ts3 and something else which happened to start with a null byte consecutively in memory. So when you tried to print ts1, the string at that address ran over ts1, ts2 and ts3 to finally stop at that null byte afterward. Having a null byte was just a stroke of luck, or maybe a stroke of unluck since it hid the bug about the missing null terminator.
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Strings in C are null-terminated. This means that a string ends with a null byte. (A byte in a string has the type char.) When you copy the bytes of a substring to a new array of bytes, you're copying the “useful” content of the string, but to get a complete string, you also need to write the null terminator. For example:
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
ts1[5] = 0;
(Don't confuse the null byte 0, which can also be written '', with the digit '0'.)
A more idiomatic way of writing this would be
memcpy(ts1, json + 32, 5);
ts1[5] = 0;
A better-designed function to copy a substring is strlcpy but unfortunately it is not part of standard C so many platforms don't have it. Note that strncpy doesn't do the job well because it wouldn't write a null byte in your case.
Note that you need 6 bytes to store a 5-byte string, i.e. you must declare char ts1[6] or char *ts1 = malloc(6) (or with a larger number).
The reason it seemed to work when you printed the strings one by one is that there must have been a null byte immediately after the string in memory. This is just a coincidence and might not happen if the function was called from different place, if the code before it had worked on different data, if compiled with a different compiler, etc. Given the output you got with the three variables, what must have happened is that the compiler happened to put ts1, ts2 and ts3 and something else which happened to start with a null byte consecutively in memory. So when you tried to print ts1, the string at that address ran over ts1, ts2 and ts3 to finally stop at that null byte afterward. Having a null byte was just a stroke of luck, or maybe a stroke of unluck since it hid the bug about the missing null terminator.
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
add a comment |
Strings in C are null-terminated. This means that a string ends with a null byte. (A byte in a string has the type char.) When you copy the bytes of a substring to a new array of bytes, you're copying the “useful” content of the string, but to get a complete string, you also need to write the null terminator. For example:
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
ts1[5] = 0;
(Don't confuse the null byte 0, which can also be written '', with the digit '0'.)
A more idiomatic way of writing this would be
memcpy(ts1, json + 32, 5);
ts1[5] = 0;
A better-designed function to copy a substring is strlcpy but unfortunately it is not part of standard C so many platforms don't have it. Note that strncpy doesn't do the job well because it wouldn't write a null byte in your case.
Note that you need 6 bytes to store a 5-byte string, i.e. you must declare char ts1[6] or char *ts1 = malloc(6) (or with a larger number).
The reason it seemed to work when you printed the strings one by one is that there must have been a null byte immediately after the string in memory. This is just a coincidence and might not happen if the function was called from different place, if the code before it had worked on different data, if compiled with a different compiler, etc. Given the output you got with the three variables, what must have happened is that the compiler happened to put ts1, ts2 and ts3 and something else which happened to start with a null byte consecutively in memory. So when you tried to print ts1, the string at that address ran over ts1, ts2 and ts3 to finally stop at that null byte afterward. Having a null byte was just a stroke of luck, or maybe a stroke of unluck since it hid the bug about the missing null terminator.
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
add a comment |
Strings in C are null-terminated. This means that a string ends with a null byte. (A byte in a string has the type char.) When you copy the bytes of a substring to a new array of bytes, you're copying the “useful” content of the string, but to get a complete string, you also need to write the null terminator. For example:
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
ts1[5] = 0;
(Don't confuse the null byte 0, which can also be written '', with the digit '0'.)
A more idiomatic way of writing this would be
memcpy(ts1, json + 32, 5);
ts1[5] = 0;
A better-designed function to copy a substring is strlcpy but unfortunately it is not part of standard C so many platforms don't have it. Note that strncpy doesn't do the job well because it wouldn't write a null byte in your case.
Note that you need 6 bytes to store a 5-byte string, i.e. you must declare char ts1[6] or char *ts1 = malloc(6) (or with a larger number).
The reason it seemed to work when you printed the strings one by one is that there must have been a null byte immediately after the string in memory. This is just a coincidence and might not happen if the function was called from different place, if the code before it had worked on different data, if compiled with a different compiler, etc. Given the output you got with the three variables, what must have happened is that the compiler happened to put ts1, ts2 and ts3 and something else which happened to start with a null byte consecutively in memory. So when you tried to print ts1, the string at that address ran over ts1, ts2 and ts3 to finally stop at that null byte afterward. Having a null byte was just a stroke of luck, or maybe a stroke of unluck since it hid the bug about the missing null terminator.
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
Strings in C are null-terminated. This means that a string ends with a null byte. (A byte in a string has the type char.) When you copy the bytes of a substring to a new array of bytes, you're copying the “useful” content of the string, but to get a complete string, you also need to write the null terminator. For example:
for(int i = 0; i<5; i++)
{
ts1[i] = json[i+32];
}
ts1[5] = 0;
(Don't confuse the null byte 0, which can also be written '', with the digit '0'.)
A more idiomatic way of writing this would be
memcpy(ts1, json + 32, 5);
ts1[5] = 0;
A better-designed function to copy a substring is strlcpy but unfortunately it is not part of standard C so many platforms don't have it. Note that strncpy doesn't do the job well because it wouldn't write a null byte in your case.
Note that you need 6 bytes to store a 5-byte string, i.e. you must declare char ts1[6] or char *ts1 = malloc(6) (or with a larger number).
The reason it seemed to work when you printed the strings one by one is that there must have been a null byte immediately after the string in memory. This is just a coincidence and might not happen if the function was called from different place, if the code before it had worked on different data, if compiled with a different compiler, etc. Given the output you got with the three variables, what must have happened is that the compiler happened to put ts1, ts2 and ts3 and something else which happened to start with a null byte consecutively in memory. So when you tried to print ts1, the string at that address ran over ts1, ts2 and ts3 to finally stop at that null byte afterward. Having a null byte was just a stroke of luck, or maybe a stroke of unluck since it hid the bug about the missing null terminator.
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
answered Nov 22 '18 at 6:57
GillesGilles
75.4k19161205
75.4k19161205
add a comment |
add a comment |
6
You aren't zero-terminating your strings.
– Lee Daniel Crocker
Nov 21 '18 at 21:36
Can you show us "printing them all at the same time" vs "printing them 1 by 1"?
– CoffeeTableEspresso
Nov 21 '18 at 21:36
Can you show the target string definitions?
– Weather Vane
Nov 21 '18 at 21:36
3
Please include a main() function and provide a Compilable, Minimal, Complete, and Verifiable Example. It makes it a lot easier to reproduce what you are seeing.
– John Murray
Nov 21 '18 at 21:43
2
"What am I doing wrong?" --> You are not posting the definitions of
ts1, ts2, ts3– chux
Nov 21 '18 at 21:55