等于在 C++ 中返回 false

2022-01-23 00:00:00 command-line-arguments g++ c++

我对 cpp 还很陌生,我正在尝试做一个项目.它说代码必须以文件名作为参数,并将由以下人员运行:

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,它将遍历参数列表以找到-i"参数,以便确定文件名.但是这一行总是返回 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,您比较了两个指针:分别为 char*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]"-i" 两个指针,而是比较不太可能指向同一位置的两个指针.因此,该检查不适用于您的情况.

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.

您可以通过多种方式修复它,例如将 "-i" 包装到 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;
    }
}

这是一种更可取的方式,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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