Passing NULL to atoi() Function results in Segmentation Fault

Gobi Sakthivel

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int main()

{

    char* x = NULL;

    int i;

    i = atoi(x);

    printf("i = %d\n",i);

    return 0;

}

This will results in Segmentation Fault.

when we pass NULL to atoi(or any functions like this) it is giving segmentation fault. My question is why they didnt check for NULL case( which should be given first priority) in such inbuilt implementations.

Banfa

Because the specification states that you will pass in a valid pointer to a string. Given that specification the check is just superfluous processing and in a language that is designed for portability and speed that is undesired.

The problem is them but rather the code that calls the function with data that is outside of the functions stated specification.

If there is a chance that the pointer is NULL the calling code should be checking for it.

swapnali143

syntax of atoi() is as follow

int atoi ( const char * str );

Because NULL is not Type Safe its giving you segmentation fault

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 // Sample program to explain atoi()

        #include<stdio.h>

    #include<stdlib.h>
 
    int main ()

    {

        int i;

        char buffer [256];
 
        printf ("Enter a number: ");

        fgets (buffer, 256, stdin);
 
        i = atoi (buffer);

        printf ("The value entered is %d.",i);
 
        return 0;

    }

Passing NULL to atoi() Function results in Segmentation Fault

✓ answered by Banfa

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