为什么sizeof（char + char）返回4？

chara,b; 
printf("%d",sizeof(a+b));

What will printf write to the screen?

printf写入屏幕的内容是什么?

I thought because sizeof(char)=1, that sizeof(a+b) will be also 1, but it turned out to be 4. I don't understand this, why does it write 4 if we are adding two chars?

我认为因为sizeof(char)= 1,sizeof(a + b)也将是1,但结果是4.我不明白这一点,如果我们添加两个字符,为什么写4?

2 个解决方案

#1

In C language operands of almost all arithmetic operators are subjected to implicit conversions called usual arithmetic conversions or, in this case, integer promotions. Operands of type char are promoted to type int and the actual addition is performed within the domain of int (or unsigned int, depending on the properties of char on that platform). So your a + b is actually interpreted as (int) a + (int) b. The result has type int and sizeof(int) is apparently 4 on your platform. That 4 is what you see.

在C语言中,几乎所有算术运算符的操作数都受到称为常规算术转换的隐式转换,或者在这种情况下是整数提升。 char类型的操作数被提升为int类型,实际的添加是在int域内执行的(或unsigned int,具体取决于该平台上char的属性)。所以你的a + b实际上被解释为(int)a +(int)b。结果的类型为int,而sizeof(int)在您的平台上显然为4。那是你看到的4。

And don't use %d to printf the result of sizeof. The result of sizeof has type size_t, while %d requires an int argument. So, either use the proper format specifier

并且不要使用%d来printf sizeof的结果。 sizeof的结果是size_t类型,而%d需要int参数。因此,要么使用正确的格式说明符

printf("%zu\n", sizeof(a+b));

or at least cast the argument if you are sure it fits

或者至少在你确定它适合的情况下抛出论据

printf("%d\n", (int) sizeof(a+b));

2 个解决方案

#1

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