strtol的正确使用

Correct usage of strtol(strtol的正确使用)

问题描述

下面的程序将字符串转换为long，但根据我的理解，它也会返回错误.我所依赖的事实是，如果 strtol 成功地将字符串转换为 long，那么 strtol 的第二个参数应该等于 NULL.当我使用 55 运行以下应用程序时，我收到以下消息.

The program below converts a string to long, but based on my understanding it also returns an error. I am relying on the fact that if strtol successfully converted string to long, then the second parameter to strtol should be equal to NULL. When I run the below application with 55, I get the following message.

./convertToLong 55
Could not convert 55 to long and leftover string is: 55 as long is 55

如何从 strtol 中成功检测错误?在我的应用程序中，零是一个有效值.

How can I successfully detect errors from strtol? In my application, zero is a valid value.

代码:

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

static long parseLong(const char * str);

int main(int argc, char ** argv)
{
    printf("%s as long is %ld
", argv[1], parseLong(argv[1]));
    return 0;
 }

static long parseLong(const char * str)
{
    long _val = 0;
    char * temp;

    _val = strtol(str, &temp, 0);

    if(temp != '')
            printf("Could not convert %s to long and leftover string is: %s", str, temp);

    return _val;
}

推荐答案

大功告成.temp 本身不会为空，但是如果整个字符串被转换，它会指向一个空字符，所以你需要解引用它:

You're almost there. temp itself will not be null, but it will point to a null character if the whole string is converted, so you need to dereference it:

if (*temp != '')

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