用C命令行参数不正确打印 [英] Command Line argument in C not printing correctly

问题描述

我尝试测试我的程序是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:

我的命令行输入测试方法是：

my command line input to test is:

"./concordance 15 < input.txt"

程序的其他工作，但测试的参数。所以在我的主要功能，我有这样的：

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，该程序打印：

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

./concordance
15
2

我的程序工作，我需要打开input.txt的文件，所以我的问题是，为什么是程序只打印./concordance和15藏汉因为只有看到2个参数，如果我有＆LT; 并在命令行input.txt的？

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?

感谢

推荐答案

有是命令行和标准输入之间的差异 - 15是一个命令行参数。外壳看到'＆LT;'和重定向的文件，你的程序在它的标准输入流。

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.

如果你不希望使用标准输入来处理文件，只需通过它的名字，并打开自己： ./一致性15 input.txt的（的argv [2]将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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