scanf produces segfault on the line of scanf
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf(n);
return 0;
}
This is the code I have, and I'm so dumbed out probably missing out on a very basic mistake. It segfaults on the line of scanf.
How would I resolve this problem?
c scanf
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
|
show 1 more comment
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf(n);
return 0;
}
This is the code I have, and I'm so dumbed out probably missing out on a very basic mistake. It segfaults on the line of scanf.
How would I resolve this problem?
c scanf
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
2
printf needs a string first. Then the parameters. likeprintf("%d",n);
– KaeptnNemo
Nov 19 at 21:42
7
No it doesn't. It segfaults onprintf(n). The first argument toprintf()must be a pointer to a format string. You're passing in the numeric value ofnas a pointer (and your compiler would almost certainly warn about this if compiled with a sufficiently sensitive warning level such as-Wallwithgcc).
– TypeIA
Nov 19 at 21:42
Thanks! @TypeIA
– Joseph Seung Jae Dollar
Nov 19 at 21:46
3
The text-book reason for downvotes on questions is "This question does not show any research effort ...". Could you demonstrate your research effort, by explaining how even while only reading the printf spec you could miss the need for more than one parameter?
– Yunnosch
Nov 19 at 21:50
1
@stark True, i.e. not strictly. But for any attempt to printnit does.
– Yunnosch
Nov 20 at 18:35
|
show 1 more comment
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf(n);
return 0;
}
This is the code I have, and I'm so dumbed out probably missing out on a very basic mistake. It segfaults on the line of scanf.
How would I resolve this problem?
c scanf
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf(n);
return 0;
}
This is the code I have, and I'm so dumbed out probably missing out on a very basic mistake. It segfaults on the line of scanf.
How would I resolve this problem?
c scanf
c scanf
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
asked Nov 19 at 21:40
Joseph Seung Jae Dollar
3822621
3822621
2
printf needs a string first. Then the parameters. likeprintf("%d",n);
– KaeptnNemo
Nov 19 at 21:42
7
No it doesn't. It segfaults onprintf(n). The first argument toprintf()must be a pointer to a format string. You're passing in the numeric value ofnas a pointer (and your compiler would almost certainly warn about this if compiled with a sufficiently sensitive warning level such as-Wallwithgcc).
– TypeIA
Nov 19 at 21:42
Thanks! @TypeIA
– Joseph Seung Jae Dollar
Nov 19 at 21:46
3
The text-book reason for downvotes on questions is "This question does not show any research effort ...". Could you demonstrate your research effort, by explaining how even while only reading the printf spec you could miss the need for more than one parameter?
– Yunnosch
Nov 19 at 21:50
1
@stark True, i.e. not strictly. But for any attempt to printnit does.
– Yunnosch
Nov 20 at 18:35
|
show 1 more comment
2
printf needs a string first. Then the parameters. likeprintf("%d",n);
– KaeptnNemo
Nov 19 at 21:42
7
No it doesn't. It segfaults onprintf(n). The first argument toprintf()must be a pointer to a format string. You're passing in the numeric value ofnas a pointer (and your compiler would almost certainly warn about this if compiled with a sufficiently sensitive warning level such as-Wallwithgcc).
– TypeIA
Nov 19 at 21:42
Thanks! @TypeIA
– Joseph Seung Jae Dollar
Nov 19 at 21:46
3
The text-book reason for downvotes on questions is "This question does not show any research effort ...". Could you demonstrate your research effort, by explaining how even while only reading the printf spec you could miss the need for more than one parameter?
– Yunnosch
Nov 19 at 21:50
1
@stark True, i.e. not strictly. But for any attempt to printnit does.
– Yunnosch
Nov 20 at 18:35
2
2
printf needs a string first. Then the parameters. like
printf("%d",n);– KaeptnNemo
Nov 19 at 21:42
printf needs a string first. Then the parameters. like
printf("%d",n);– KaeptnNemo
Nov 19 at 21:42
7
7
No it doesn't. It segfaults on
printf(n). The first argument to printf() must be a pointer to a format string. You're passing in the numeric value of n as a pointer (and your compiler would almost certainly warn about this if compiled with a sufficiently sensitive warning level such as -Wall with gcc).– TypeIA
Nov 19 at 21:42
No it doesn't. It segfaults on
printf(n). The first argument to printf() must be a pointer to a format string. You're passing in the numeric value of n as a pointer (and your compiler would almost certainly warn about this if compiled with a sufficiently sensitive warning level such as -Wall with gcc).– TypeIA
Nov 19 at 21:42
Thanks! @TypeIA
– Joseph Seung Jae Dollar
Nov 19 at 21:46
Thanks! @TypeIA
– Joseph Seung Jae Dollar
Nov 19 at 21:46
3
3
The text-book reason for downvotes on questions is "This question does not show any research effort ...". Could you demonstrate your research effort, by explaining how even while only reading the printf spec you could miss the need for more than one parameter?
– Yunnosch
Nov 19 at 21:50
The text-book reason for downvotes on questions is "This question does not show any research effort ...". Could you demonstrate your research effort, by explaining how even while only reading the printf spec you could miss the need for more than one parameter?
– Yunnosch
Nov 19 at 21:50
1
1
@stark True, i.e. not strictly. But for any attempt to print
n it does.– Yunnosch
Nov 20 at 18:35
@stark True, i.e. not strictly. But for any attempt to print
n it does.– Yunnosch
Nov 20 at 18:35
|
show 1 more comment
2 Answers
2
active
oldest
votes
Use of printf() like bellow :
printf("%d", n);
Probably you did a mistake in your code before using scanf. For example it may be occurred if you copy a string in a uninitialized array that pointing nowhere in memory.
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
add a comment |
The segmentation fault is not at the scanf, but at the printf. As @TypeIA has mentioned in the comments section, this is because the printf expects to have a pointer to the format string and not the integer itself. To do this, just how you use the format specifier %d while accepting an integer as input through scanf, you need to use the same while printing using printf., i.e,
printf("%d",n); Notice the %d before n.
UPDATED CODE:
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf("%d",n);
return 0;
}
answered Nov 20 at 5:29
Rai
776218
1
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use of printf() like bellow :
printf("%d", n);
Probably you did a mistake in your code before using scanf. For example it may be occurred if you copy a string in a uninitialized array that pointing nowhere in memory.
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
add a comment |
Use of printf() like bellow :
printf("%d", n);
Probably you did a mistake in your code before using scanf. For example it may be occurred if you copy a string in a uninitialized array that pointing nowhere in memory.
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
add a comment |
Use of printf() like bellow :
printf("%d", n);
Probably you did a mistake in your code before using scanf. For example it may be occurred if you copy a string in a uninitialized array that pointing nowhere in memory.
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
Use of printf() like bellow :
printf("%d", n);
Probably you did a mistake in your code before using scanf. For example it may be occurred if you copy a string in a uninitialized array that pointing nowhere in memory.
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
edited Nov 20 at 5:40
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
answered Nov 20 at 5:33
Mohammadreza Panahi
2,52221432
2,52221432
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
add a comment |
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
add a comment |
The segmentation fault is not at the scanf, but at the printf. As @TypeIA has mentioned in the comments section, this is because the printf expects to have a pointer to the format string and not the integer itself. To do this, just how you use the format specifier %d while accepting an integer as input through scanf, you need to use the same while printing using printf., i.e,
printf("%d",n); Notice the %d before n.
UPDATED CODE:
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf("%d",n);
return 0;
}
answered Nov 20 at 5:29
Rai
776218
1
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
add a comment |
The segmentation fault is not at the scanf, but at the printf. As @TypeIA has mentioned in the comments section, this is because the printf expects to have a pointer to the format string and not the integer itself. To do this, just how you use the format specifier %d while accepting an integer as input through scanf, you need to use the same while printing using printf., i.e,
printf("%d",n); Notice the %d before n.
UPDATED CODE:
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf("%d",n);
return 0;
}
answered Nov 20 at 5:29
Rai
776218
1
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
add a comment |
The segmentation fault is not at the scanf, but at the printf. As @TypeIA has mentioned in the comments section, this is because the printf expects to have a pointer to the format string and not the integer itself. To do this, just how you use the format specifier %d while accepting an integer as input through scanf, you need to use the same while printing using printf., i.e,
printf("%d",n); Notice the %d before n.
UPDATED CODE:
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf("%d",n);
return 0;
}
answered Nov 20 at 5:29
Rai
776218
The segmentation fault is not at the scanf, but at the printf. As @TypeIA has mentioned in the comments section, this is because the printf expects to have a pointer to the format string and not the integer itself. To do this, just how you use the format specifier %d while accepting an integer as input through scanf, you need to use the same while printing using printf., i.e,
printf("%d",n); Notice the %d before n.
UPDATED CODE:
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
printf("%d",n);
return 0;
}
answered Nov 20 at 5:29
Rai
776218
edited Nov 23 at 4:26
answered Nov 20 at 5:29
Rai
776218
answered Nov 20 at 5:29
Rai
776218
answered Nov 20 at 5:29
Rai
776218
776218
1
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
add a comment |
1
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
1
1
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
This answer lacks explanation for the segfault. A good one is already described on this page.
– Yunnosch
Nov 20 at 18:37
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
Thanks@Yunnosch, I have updated the answer with a line of explanation.
– Rai
Nov 21 at 4:19
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
You could probably improve the answer by using the better, more detailed explanation in the comments, giving credits of course and maybe even asking for permission.
– Yunnosch
Nov 21 at 11:49
add a comment |
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2
printf needs a string first. Then the parameters. like
printf("%d",n);– KaeptnNemo
Nov 19 at 21:42
7
No it doesn't. It segfaults on
printf(n). The first argument toprintf()must be a pointer to a format string. You're passing in the numeric value ofnas a pointer (and your compiler would almost certainly warn about this if compiled with a sufficiently sensitive warning level such as-Wallwithgcc).– TypeIA
Nov 19 at 21:42
Thanks! @TypeIA
– Joseph Seung Jae Dollar
Nov 19 at 21:46
3
The text-book reason for downvotes on questions is "This question does not show any research effort ...". Could you demonstrate your research effort, by explaining how even while only reading the printf spec you could miss the need for more than one parameter?
– Yunnosch
Nov 19 at 21:50
1
@stark True, i.e. not strictly. But for any attempt to print
nit does.– Yunnosch
Nov 20 at 18:35