Char 没有转换为 int [英] Char isn't converting to int

问题描述

出于某种原因，我的 C 程序拒绝将 argv 的元素转换为整数，我不知道为什么.

For some reason my C program is refusing to convert elements of argv into ints, and I can't figure out why.

int main(int argc, char *argv[])
{

    fprintf(stdout, "%s\n", argv[1]);

    //Make conversions to int
    int bufferquesize = (int)argv[1] - '0';

    fprintf(stdout, "%d\n", bufferquesize);
}

这是运行 ./test 50 时的输出:

And this is the output when running ./test 50:

50

-1076276207

-1076276207

我尝试删除 (int)，同时抛出 * 和 &(int) 和 argv[1] 之间 - 前者给了我 5 但不是 50，但后者给了我类似于上面的输出.删除 - '0' 操作无济于事.我还尝试先创建一个 char = argv[1] 并使用 first 进行转换，这很奇怪，无论输入如何，我都得到了 17.

I have tried removing the (int), throwing both a * and an & between (int) and argv[1] - the former gave me a 5 but not 50, but the latter gave me an output similar to the one above. Removing the - '0' operation doesn't help much. I also tried making a char first = argv[1] and using first for the conversion instead, and this weirdly enough gave me a 17 regardless of input.

我非常困惑.这是怎么回事?

I'm extremely confused. What is going on?

推荐答案

argv[1] is a char * not a char you无法将 char * 转换为 int.如果您想将 argv[1] 中的第一个字符更改为 int，您可以这样做.

argv[1] is a char * not a char you can't convert a char * to an int. If you want to change the first character in argv[1] to an int you can do.

int i = (int)(argv[1][0] - '0');

我刚写的

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

int main(int argc, char **argv) {
    printf("%s\n", argv[1]);

    int i = (int)(argv[1][0] - '0');

    printf("%d\n", i);
    return 0;
}

然后像这样运行

./testargv 1243

得到

1243
1

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