Why in the code “456”+1, output is “56”

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于 2015-03-26 08:04:50 发布

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本文探讨了在C++中使用字符串字面量进行数值拼接时遇到的问题，并解释了为何输出结果不符合预期的原因。文章详细说明了字符串字面量在内存中的表示方式及其与指针操作之间的关系。

Question:

#include <iostream>
int main()
{
    std::cout << "25"+1;
    return 0;
}
I am getting "5" as output.when I use "5"+1,output is blank;"456"+1 output is "56".confused what is going behind the scenes.


Answer:
The string literal "25" is really a char array of type const char[3] with values {'2', '5', '\0'} (the two characters you see and a null-terminator.) In C and C++, arrays can easily decay to pointers to their first element. This is what happens in this expression:
"25" + 1

where "25" decays to &"25"[0], or a pointer to the first character. Adding 1 to that gives you a pointer to 5.

On top of that, std::ostream, of which std::cout is an instance, prints a const char* (note that char* would also work) by assuming it is a null-terminated string. So in this case, it prints only 5.