strtol reusing param

问题:

This code seems to work as expected, populates an array of numbers using a single pointer

#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>

int main(void)
{
    int arr[4], count = 0, i;
    char *p, s[32] = "  \t  10,  15  \n  ,20,   25  , ";

    p = s;
    do {
        arr[count++] = (int)strtol(p, &p, 10);
        while (isspace(*p) || *p == ',') p++;
    } while (*p);
    for (i = 0; i < count; i++) {
        printf("%d\n", arr[i]);
    }
    return 0;
}

My question is:

It is valid to use p as param1 (source) and &p as param 2 (address of the first invalid character) in strtol?

回答1:

Yes it is safe.

Please refer to http://en.cppreference.com/w/cpp/string/byte/strtol for complete usage reference. Line 11 of the example illustrates a call using the same variable for the 1st and 2nd parameters.

回答2:

Yes, it is safe. The first argument is passed by value, so strtol has a local copy that isn't affected by changes written to the second parameter.

回答3:

Yes, this is valid, as you are keeping the pointer to the beginning of the string (pointer s). Consider that you have this situation:

#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    int arr[4], count = 0, i;
    char *p, *s;
    s = (char*)malloc(sizeof(char) * 15);
    strcpy(s, "  \t  10,  15  \n  ,20,   25  , ");

    p = s;
    do {
        arr[count++] = (int)strtol(p, &p, 10);
        while (isspace(*p) || *p == ',') p++;
    } while (*p);
    for (i = 0; i < count; i++) {
        printf("%d\n", arr[i]);
    }
    free(s);
    return 0;
}

strtol will move p pointer to somewhere in string. If you call free(p) you will have memory leak (if it doesn't fail). But, since you are keeping s pointer, you will always be able to free the occupied memory.