strtol函数的用法——字符串转长整形

/* strtol example */

#include <stdio.h>      /* printf */

#include <stdlib.h>     /* strtol */

int main ()

{

    char szNumbers[] = "2001 60c0c0 -1101110100110100100000 0x6fffff";

    char *pEnd;

    long int li1, li2, li3, li4;

    li1 = strtol (szNumbers, &pEnd, 10);

    printf("pEnd = %s\n", pEnd);

    li2 = strtol (pEnd, &pEnd, 16);

    printf("pEnd = %s\n", pEnd);

    li3 = strtol (pEnd, &pEnd, 2);

    printf("pEnd = %s\n", pEnd);

    li4 = strtol (pEnd, NULL, 0);

    printf ("The decimal equivalents are: %ld, %ld, %ld and %ld.\n", li1, li2, li3, li4);

    return 0;

}

运行结果：

pEnd = 60c0c0 -1101110100110100100000 0x6fffff

pEnd = -1101110100110100100000 0x6fffff

pEnd = 0x6fffff

The decimal equivalents are: 2001, 6340800, -3624224 and 7340031.

从结果中可以看出，pEnd是转换后剩余的字符串

巴特西

strtol函数的用法——字符串转长整形

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