为什么main()参数argv类型为char * []而不是const char * []?

问题描述：

当我编写并执行以下代码时，编译器说

When I wrote the following code and executed it, the compiler said

不建议从字符串常量转换为char*

int main()  
{  
  char *p;  
  p=new char[5];  
  p="how are you";  
  cout<< p;  
  return 0;  
}

这意味着我应该写const char *.

但是，当我们使用char* argv[]将参数传递给main时，我们不要编写const char* argv[].

But when we pass arguments into main using char* argv[] we don't write const char* argv[].

为什么?

答

因为... argv[]不是const.而且它肯定不是(静态)字符串文字，因为它是在运行时创建的.

Because ... argv[] isn't const. And it certainly isn't a (static) string literal since it's being created at runtime.

您要声明一个char *指针，然后为其分配一个字符串文字，根据定义，该文字是常量；实际数据位于只读存储器中.

You're declaring a char * pointer then assigning a string literal to it, which is by definition constant; the actual data is in read-only memory.

int main(int argc, char **argv)  {
    // Yes, I know I'm not checking anything - just a demo
    argv[1][0] = 'f';
    std::cout << argv[1] << std::endl;
}

输入:

g ++ -o测试test.cc

g++ -o test test.cc

./test hoo

输出:

foo

这不是您要更改argv的为什么的评论，但肯定有可能.

This is not a comment on why you'd want to change argv, but it certainly is possible.

为什么main()参数argv类型为char * []而不是const char * []?

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