为什么`equal'用于C ++中的const char *？ [英] Why does the `equal` works for const char* in C++?

问题描述

代码如下，它输出 1 ：

  int main（int argc，char * argv []）
 {
 vector< const char *> vec = {nima，123}; 
 vector< const char *> vec2 = {nima，123}; 
 auto result = equal（vec.cbegin（），vec.cend（），vec2.cbegin（））; 
 cout<<结果< endl; 
 return 0; 
} 
  

我知道我可以测试两个c风格的字符串是否相等使用 strcmp （因为 char * 不是我理解的对象）。但是等于是< algorithm> 的函数。它是否重载 == 运算符，以便它可以测试两个 char * ？ b
$ b

@ Damon说，它们等于将同一个字符串文字合并到同一个地址，就像我的理解。但是，当我尝试不同地址的 char * 时，它仍然给出了相同的结果：

  int main（int argc，char * argv []）
 {
 char * k =123 
 char * b =123; 
 vector< const char *> vec = {nima}; 
 vector< const char *> vec2 = {nima}; 
 cout<< & k< endl; 
 cout<< & b< endl; 
 vec.push_back（k）; 
 vec2.push_back（b）; 
 
 auto result = equal（vec.cbegin（），vec.cend（），vec2.cbegin（））; 
 cout<<结果< endl; 
 return 0; 
} 
  

结果是：

  0x7fff5f73b370 
 0x7fff5f73b368 
 1 
  

解决方案

这里可能发生的是编译器/链接器将其中两个相同的四个字符串文字合并为两个文字。因此，nima和123都有相同的地址。

您将地址存储在向量中并将它们（地址上的 operator == ）进行比较。由于地址相同，比较是相等的。

请注意，这是 accidential 。它只有工作，因为两个原因：

1. 字符串是字面量的（也就是说，不是从<$ c $


2. 允许编译器，但不需要合并相同的字面值（无论这是否是实现定义的）。

这可能会导致两种情况，你会得到一个有趣的惊喜（如果你必须找出为什么突然不工作，它工作所有的时间），当你使用不同的编译器，甚至相同的编译器，不同的优化设置，或当你分配非文字字符串。

The codes are like this, it outputs 1:

int main(int argc, char *argv[])
{
    vector <const char*> vec = {"nima","123"};
    vector <const char*> vec2 = {"nima","123"};
    auto result = equal(vec.cbegin(), vec.cend(), vec2.cbegin());
    cout << result << endl;
    return 0;
}

I knew that I can test whether two c-style string is equal only by using strcmp (because char* is not an object as I understood). But here equal is a function from <algorithm>. Does it overload the == operator so that it can test the equality of two char*?

@Damon says that they're equal as the merges the same string literal into same address, as I understood. However, when I tried char* with different addresses, it still gives me the same result:

int main(int argc, char *argv[])
{
char* k = "123";
char* b = "123";
vector <const char*> vec = {"nima"};
vector <const char*> vec2 = {"nima"};
cout << &k << endl;
cout << &b << endl;
vec.push_back(k);
vec2.push_back(b);

auto result = equal(vec.cbegin(), vec.cend(), vec2.cbegin());
cout << result << endl;
return 0;
}

The result is:

 0x7fff5f73b370
 0x7fff5f73b368
 1

解决方案

What probably happens here is that the compiler/linker merges the four string literals of which two are identical into two literals. Therefore, both "nima" and both "123" have the same address.

You store the addresses in a vector and compare them (operator== on an address). Since the addresses are the same, the comparison is equal.

Note that this is accidential. It only "works" because of two reasons:

1. The strings are literals (that is, not some strings read from e.g. stdin).
2. A compiler is allowed, but not required to merge identical literals (whether or not this happens is implementation-defined).

This can lead to two situations in which you get a funny surprise (not so funny if you must find out why it suddenly doesn't work when it "worked" all the time), either when you use a different compiler or even the same compiler with different optimization settings, or when you assign non-literal strings.

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