字符串文字是否在C编译时自动转换为char *? [英] Are string literals automatically being casted to char* at compile time in C?

问题描述

如果我要执行以下操作:

If I were to do something like:

printf("The string is: %s\n", "string1");

在编译时是否已完成以下操作:

Is the following done at compile time:

printf("The string is: %s\n", (unsigned char*) "string1"); 

还是类似的?

推荐答案

标准定义字符串文字的类型为char ¹ 的数组，并且该数组自动衰减为指针，即char*.使用%s说明符时，无需将其作为参数传递给printf时显式进行强制转换.

It is defined by the standard that the type of string literals is an array of char¹ and arrays automatically decay to pointers, i.e. char*. You don't need to cast it explicitly while passing it as an argument to printf when %s specifier is used.

侧面说明:在C ++中为const char* ² .

Side note: In C++ it's const char*².

^[1] C99 6.4.5:字符串文字是由零个或多个多字节字符括起来的序列， 双引号，例如"xyz" ... 静态存储持续时间和长度的数组 足以包含序列.对于字符串文字，数组元素具有 输入char"

^[1] C99 6.4.5: "A character string literal is a sequence of zero or more multibyte characters enclosed in double-quotes, as in "xyz"... an array of static storage duration and length just sufficient to contain the sequence. For character string literals, the array elements have type char"

^[2] C ++ 03 2.13.4§1:"普通字符串文字的类型为"n个数组 const char"和静态存储持续时间"

^[2] C++03 2.13.4 §1: "an ordinary string literal has type "array of n const char" and static storage duration"

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