Sizeof字符串字面量

问题描述

以下代码

  #include< iostream> 
 using namespace std; 
 
 int main（）
 {
 const char * const foo =f; 
 const char bar [] =b; 
 cout<< sizeof（string literal）=< sizeof（f）<< endl; 
 cout<< sizeof（const char * const）=< sizeof（foo）< endl; 
 cout<< sizeof（const char []）=< sizeof（bar）< endl; 
} 
  

输出

  sizeof（string literal）= 2 
 sizeof（const char * const）= 4 
 sizeof（const char []）= 2 
 
 
 <$> 
 $ b> 
 li>为什么 sizeof 计算字符串文字的长度（所需的空间） 
 
字符串文字有不同的当给予 sizeof ？
 
 
 
 
解决方案
 
 
  sizeof（f）必须返回2，一个用于'f'在32位计算机上返回4，在64位计算机上返回8。
   
 
  sizeof（bar）返回2，因为bar是两个字符的数组，'b'和
 
 
字符串文字的类型为'char'的大小N的数组，其中N包括
 
 
 
请记住，数组传递给 sizeof 时不会衰减指针。
 
The following code
#include <iostream>
using namespace std;

int main()
{
    const char* const foo = "f";
    const char bar[] = "b";
    cout << "sizeof(string literal) = " << sizeof( "f" ) << endl;
    cout << "sizeof(const char* const) = " << sizeof( foo ) << endl;
    cout << "sizeof(const char[]) = " << sizeof( bar ) << endl;
}
outputs
sizeof(string literal) = 2
sizeof(const char* const) = 4
sizeof(const char[]) = 2
on a 32bit OS, compiled with GCC.
Why does sizeof calculate the length of (the space needed for) the string literal ?
Does the string literal have a different type (from char* or char[]) when given to sizeof ?
 解决方案 
sizeof("f") must return 2, one for the 'f' and one for the terminating '\0'.
sizeof(foo) returns 4 on a 32-bit machine and 8 on a 64-bit machine because foo is a pointer
sizeof(bar) returns 2 because bar is an array of two characters, the 'b' and the terminating '\0'.
The string literal has the type 'array of size N of char' where N includes the terminal null.
Remember, arrays do not decay to pointers when passed to sizeof.
                        
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