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

问题描述

出于某种原因，我的C程序拒绝的argv元素转换成整数，我想不通为什么。

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);
}

和运行./test 50时，这是输出：

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

50

-1076276207

-1076276207

我曾尝试删除（INT），抛既是*和＆安培;之间（INT）和argv [1] - 前者给了我一个5而不是50，但后者给了我类似上面的那个的输出。拆除 - 0的操作没有太大的帮助。我也试过做一个char第一=的argv [1]，并首次使用了转换，而是和这个古怪足够给了我一个17无论输入的。

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

的argv [1] 是的char * 不是字符您不能在的char * 转换为 INT 。如果你想argv中[1]的第一个字符更改为int你可以做。

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');

我刚写了这个

#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;
}

和运行它像这样

./testargv 1243

和获得

1243
1

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