Why is the second argument not working with strtol? - c

Question

I did this something like this:
/* convert the argv[1] into Integer and store the result in key
* using library function: strtol() to do it */
char **flag = NULL;
key = strtol(argv[1], flag, 10);
// if argv[1] is not all digits
if (**flag != '\0')
{
printf("Usage: ./caesar key\n");
return 1;
}
But it throws a segmentation fault. I am not sure why.
In the C documentation, strtol is prototyped as long int strtol(const char *nptr, char **endptr, int base). Why am I getting a segmentation fault?
And when change some parts of the code to char *flag, strtol(argv[1], &flag, 10), and if (*flag != '\0'), everything works as expected.
I understand (sort of) how making the edits corrects the code. However, I do not know why the original code does not work. Does anyone have a clue?

Answer 1

You need to pass the address of your char * pointer, so that strtol can update it.
char *flag = NULL;
key = strtol(argv[1], &flag, 10);
// if argv[1] is not all digits
if (*flag != '\0')
After the call, the pointer will be modified to point to the end of the parsed region of the input string.
The man page wording is admittedly confusing here.

Answer 2

I do not know why the original code does not work? Does anyone have any clue?
Because here
char **flag = NULL;
flag is set to NULL, then here
key = strtol(argv[1], flag, 10);
flags's value (NULL) is passed to strtol(), which does not change flags's value in any way, it is NULL before and after the call to strtol().
and finally here
if (**flag != '\0')
flag, which has the value NULL, is dereferenced, which invokes undefined behaviour, which could lead to anything, which in your case is a crash.

Answer 3

Consider this simple function which multiplies a number by 2.
void Times2(int number, int *result)
{
*result = number * 2;
}
You should call it like this:
int result;
Times2(3, &result);
printf("%d\n", result);
and the expected output would be 4.
But if you call it like this (which is basically what you're doing when you call strtol in your buggy code):
int *result = NULL;
Times2(3, result);
printf("%d\n", result);
your program will most likely end up with a segmentation fault, because in the Times2 function you dereference a NULL pointer.

Answer 4

Why second argument not working with strol?
char **flag = NULL;
key = strtol(argv[1], flag, 10);
It is working. Passing is a null pointer is fine, but flag will not get changed in the calling code.
When the 2nd argument is a null pointer, nothing is updated about where the conversion ended.
Later code: *flag in if (**flag != '\0') is bad (undefined behavior) as that attempts to de-reference a null pointer.
Instead pass the address of a known pointer:
//char **flag = NULL;
char *flag;
//key = strtol(argv[1], flag, 10);
key = strtol(argv[1], &flag, 10);
// if (**flag != '\0')
if (*flag != '\0')
Following is a complete test conversion of a char* to an int.
#include <stdlib.h>
#incluse <ctype.h>
// Return error flag
int my_strtol(long *destination, const char *s) {
if (destination == NULL) {
return 1;
}
if (s == NULL) {
*destination = 0;
return 1;
}
char *endptr;
errno = 0;
*destination = strtol(s, &endptr, 0);
// If no conversion or out-of-range
if (s == endptr || errno == ERANGE) {
return 1;
}
// I like to tolerate trailing white-space.
while (isspace(*(const unsigned char *) endptr)) {
endptr++;
}
return *endptr != '\0';
}