C 中的命令行参数打印不正确 [英] Command Line argument in C not printing correctly

问题描述

我正在尝试测试我的程序如何接收用户命令行输入:

im trying to test how my program is receiving a users command line input:

我要测试的命令行输入是:

my command line input to test is:

"./concordance 15 < input.txt"

程序的其余部分工作，但测试参数.所以在我的主要功能中，我有这个:

the rest of the program works but to test the arguments. so in my main function i have this:

int main(int argc, char *argv[])
{
    int i;
    for (i = 0; i < argc; i++)
    {
        printf("%s\n", argv[i]);          //runs through command line for arg
    }
    printf("%d\n", argc);                 //prints total arguments
    return 0;
}

问题是当我输入命令行时，程序打印:

The problem is when I enter my command line, the program prints:

./concordance
15
2

为了让我的程序工作，我需要打开 input.txt 文件，所以我的问题是，为什么程序只打印./concordance"和15"，并且如果我有<"和命令行中的input.txt"?

for my program to work I need to open the input.txt file so my question is, why is the program only printing "./concordance", and "15" aswell as only seeing 2 arguments if I have "<" and "input.txt" in the command line?

谢谢

推荐答案

命令行和stdin"之间有一个区别——15"是一个命令行参数.外壳看到<"并将文件重定向"到您的程序中的 stdin 流.

There is a difference between the command line and "stdin" - "15" is a command line argument. The shell sees the '<' and "redirects" the file to your program on it's stdin stream.

如果您不想使用 stdin 来处理文件，只需传递它的名称并自己打开:./concordance 15 input.txt(argv[2] 将是 input.txt)

If you don't want to use stdin to process the file, just pass it's name and open yourself: ./concordance 15 input.txt (argv[2] will be input.txt)

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