Welcome to Software Development on Codidact!
Will you help us build our independent community of developers helping developers? We're small and trying to grow. We welcome questions about all aspects of software development, from design to code to QA and more. Got questions? Got answers? Got code you'd like someone to review? Please join us.
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);
}
1 answer
+3
−0
Bugs:
-
int **range = 0;should beint *range;and then call the function likeft_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, becauseikeeps increasing and eventually you accessrangeout 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
intthe nameptr... -
lentghtypo ->length
0 comment threads
3 comment threads