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C语言中按%d打印char会不会把相邻内存的也print出来? 第1页

  

user avatar   gou-rou-hua-yu-guo 网友的相关建议: 
      

不会,因为C标准是这么规定的

6.5.2.2:6
If the expression that denotes the called function has a type that does not include a
prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions. .......
— one promoted type is a signed integer type, the other promoted type is the corresponding unsigned integer type, and the value is representable in both types;
— both types are pointers to qualified or unqualified versions of a character type or void


6.5.2.2:7
If the expression that denotes the called function has a type that does include a prototype, the arguments are implicitly converted, as if by assignment, to the types of the corresponding parameters, taking the type of each parameter to be the unqualified version of its declared type. The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.

简单说呢,根据6.5.2.2:7 可变参数要做default argument promotions

那么什么叫default argument promotions呢?在6.5.2.2:6里规定了,整型做integer promotions, float则转换至double就叫default argument promotions

至于什么叫integer promotions呢?

6.3.1.1:2
If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.

简单说,如果int存得下,那就转换成int,否则转换成unsigned int,这叫integer promotions

所以答案是,变参里的char类型参数会被转换成int类型


PS:

这里顺便纠正一个广为流传的关于printf格式控制符的错误:

很多人提到printf用法时,都会说%f打印的float,%lf打印double,这是错误的

根据上面提到的C标准,我们可以知道变参里的float类型变量会被转换成double类型变量,因此printf不可能接收float类型的变量, %f打印的是double




  

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