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Q&A

Why does this code that uses a pointer-to-pointer-to-int segfault?

+1
−3

Hello folks, can someone resolve this seg fault with me please, i can't find the error where it occur. Thank you.

#include <stdlib.h>
 #include <stdio.h>
 int ft_ultimate_range(int **range, int min, int max)
 {
    int lentgh;
    int i;
    int mi;
    
    mi = min;
    lentgh = max - min;
    if (min >= max)
    {
        // *range = NULL;
        return (0);
    }
    *range = malloc(sizeof(int) * lentgh);
    if(!range)
    {
        return 0;
    }
    i = 0;
    while (min < max)
    {
        range[0][i] = mi;
        i++;
        mi++;
    }
    return (i);
}

int main()
 {
    int i = 0;
    int min = 5;
    int max = 19;
    int **range = 0;
    int ptr = ft_ultimate_range(range, min, max);`

    while(i < max - min)
    {
        printf("%d\n", ptr);
        i++;
    }
    return (0);
}
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3 comment threads

Infinite loop (3 comments)
when i compile it i got no error, but when i run it using ./a.out i got : "segmentation fault ./a.ou... (1 comment)
What does the debugger tell you? (1 comment)

1 answer

+3
−0

Bugs:

  • int **range = 0; should be int *range; and then call the function like ft_ultimate_range(&range, min, max);. That's the whole reason the parameter is pointer-to-pointer, so that you can return a malloc:ed pointer through a parameter.
  • while (min < max) neither min nor max change inside the loop. This is the reason for the seg fault, because i keeps increasing and eventually you access range out of bounds.
  • while(i < max - min) in main() doesn't make any sense, you can simply return that loop. You probably intended to do something with the allocated data instead here?

Minor stuff:

  • There's no obvious reason why you can't simply return the pointer from the function. It then becomes int* ft_ultimate_range(int min, int max) and all the pointer-to-pointer syntax can be removed.
  • Don't give a plain int the name ptr...
  • lentgh typo -> length
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