Why do I get this Segmentation Fault?

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Why when I declare:

char *s = "string";

fscanf(stdin, "%s", s);

I get a Segmentation Fault?

c

  • You’re trying to record something on Stdin; wouldn’t be for Stdout that you should send the string?

1 answer

5

The code below has the following problems:

char *s = "string";
sprintf(stdin, "%s", s);
  • stdin is the type FILE *, and the first parameter of sprintf is a variable of the type char *. Do not mix the types, or the result is undefined. To write to FILE *, use fprintf.
  • You can read from entree standard; if you want write down, you must use the exit pattern.

The code below does what (I think) you want.

void main() {
    char *s = "string";
    fprintf(stdout, "%s", s);
}

[More details after the question is edited] To the code below:

char *s = "string";
fscanf(stdin, "%s", s);

When you declare a literal string in a C program, the compiler usually (*) defines these literals at memory addresses that are read-only; then if you try to modify it, the program will raise a Segmentation fault. If you want to avoid this, you can either declare the variable in the stack (e. g., char s[20] = "string";) or in the heap (e. g., char *s = (char *)malloc(20 * sizeof(char)); strcpy(s, "string");. (**)

(*) This is platform dependent; on some platforms / compilers, your code may work smoothly, but you should not count on it.

(**) The use of scanf to string functions is dangerous, consider using functions that limit the number of characters read so that there is no danger of buffer overflows.

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