Segmentation fault in C

Question

void main() {
    int num;
    num = 1;
    num++;
    num = num + 9;
    printf('%u',num);
}

What is wrong with this? It says segmentation fault. Written in C.

Although in this case it is fairly obvious which line the error is on, please include this info explicitly in your future posts - it makes everyone's life easier (including your own). — Péter Török, Apr 04 '11 at 14:35
It's a good idea to look at the output of your compiler. I am sure it did warn you already about the printf(). — Mackie Messer, Apr 04 '11 at 14:42

score 8 · Answer 1 · edited May 23 '17 at 11:47

8

'%u' should be "%u" (double quotes). C is not SQL, and '%u' is considered a multicharacter literal.

And as Erik said, %d is better in your case.

edited May 23 '17 at 11:47

Community

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answered Apr 04 '11 at 14:27

Benoit

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@Ian - If this solved your issue, you should mark this answer as correct. – Varun Madiath Apr 04 '11 at 17:06

score 3 · Answer 2 · answered Apr 04 '11 at 14:28

3

printf("%d") - printf takes a C string not a character. %d is the correct format specifier for an integer.

answered Apr 04 '11 at 14:28

Erik

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score 3 · Answer 3 · answered Apr 04 '11 at 14:28

3

That '%u' ( a multi-character char constant) should be "%u" (a char array). The char constant is being misinterpreted as a pointer to random memory.

answered Apr 04 '11 at 14:28

Ernest Friedman-Hill

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score 0 · Answer 4 · answered Apr 04 '11 at 17:03

Note that you would have gotten an error from passing the wrong type (int instead of const char *) as the first argument of printf, except that you forgot to #include <stdio.h> or prototype printf yourself. And since printf is a variadic function, this results in UB too.

Segmentation fault in C

4 Answers4