等于在C ++中返回false [英] Equals returning false in c++

问题描述

我对cpp还是很陌生，我正在尝试做一个项目。它说代码必须以文件名作为参数，并将由以下命令运行：

I'm fairly new to cpp and I am trying to do a project. It says that the code must take in a filename as an argument and will be run by:

./main -i filename

我编写了一个for循环，该循环将遍历参数列表以查找 -i参数，以便我可以确定文件名。但是此行始终返回false：

I have written a for-loop that will iterate through the list of arguments to find the "-i" argument so that I can determine the filename. But this line always return false:

argv[i] == "-i"

下面是我的代码：

#include <string>
#include <iostream>

int main(int argc, char *argv[]) {
    std::string test = argv[0];
    for(int i = 0; i < argc; i++){
        if(argv[i] == "-i"){
            test = argv[i+1];
            break;
        }
    }
    std::cout << test;
    return 1;
}

推荐答案

argv[i] == "-i"

在上面的行中比较两个指针：分别为 char * 和 const char * 。

In the line above you compare two pointers: char* and const char*, respectively.

换句话说，不是比较 argv [i] 和-i 两个指针，而是比较指向同一位置的可能性很小。结果，该支票在您的情况下不起作用。

In other words, instead of comparing argv[i] and "-i" two pointers are compared which are pretty much unlikely to point to the same location. As a result, the check doesn't work in your case.

您可以采用多种方式来修复它，例如，包装-i 放入 std :: string 以使比较正常工作：

You can fix it in multiple ways, for example wrap "-i" into std::string to make the comparison work properly:

const auto arg = std::string{ "-i" };

for(int i = 0; i < argc; i++){
    if(argv[i] == arg){
        test = argv[i+1];
        break;
    }
}

从 C ++ 17开始，您还可以使用 std :: string_view ：

Starting with C++17 you might also use a std::string_view:

const std::string_view sv{ "-i" };

for(int i = 0; i < argc; i++){
    if(argv[i] == sv){
        test = argv[i+1];
        break;
    }
}

这是一种更好的方法，因为它避免了 std :: string 创建。

which is a preferable way as it avoids a std::string creation.

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