int如何将类型提升为char？ [英] How does an int gets type promoted to a char ?

问题描述

展开 | 选择 | Wrap | 行号

解决方案

你应该做的第一件事就是搞错了

展开 | 选择 | Wrap | 行号

没有促销到char的东西。


促销意味着将较小的类型变为更大的类型。就像将char（1个字节）提升为int（4个字节）一样。


％c和％d是printf格式说明符，您可以在其中请求将数据显示为char或作为一个int。


在这种情况下，数组的int由printf强制转换为char。很可能int的最低有效字节（LSB）成为char。 int的剩余字节都会丢失。

这就是我想说的0,2,3这些数字是4字节值，因此这些将被转换从LSB开始到1个字节，因此对于arr [1] = 2，op必须是与ascii值2对应的某个字符，但它不打印任何内容，虽然它为ascii值3和4输出了一些字符，那么呢？ / BLOCKQUOTE>

Expand|Select|Wrap|Line Numbers

解决方案

First thing you should do is get the errors out

Expand|Select|Wrap|Line Numbers

There is no such thing as promotion to a char.

Promotion means to make a smaller type into a bigger type. Like promotion a char (1 byte) to an int (4 bytes).

The %c and %d are printf format specifiers where you request the data to be displayed as a char or as an int.

In this case the int of the array is typecast by printf to a char. Most likely the least significant byte (LSB) of the int becomes the char. The remaining bytes of the int are lost.

So that''s what I am trying to say that 0 , 2,3 these numbers are 4 byte value , hence these would be getting converted to 1 byte starting from LSB , so for arr[1]= 2 , op must be some character corresponding to ascii value 2 , but it is not printing anything , although it prints some character for ascii value 3 and 4 , so then ?

这篇关于int如何将类型提升为char？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！