什么是分配一个无符号的变量有符号值的交易？ [英] What is the deal with assigning an unsigned variable to a signed value?

问题描述

这code，我看有很多，我看到这样的事情发生的地方：结果

This code I am looking at has a lot of places where I see things like this happening:

char *functionName(char *passedVariable)
{
    unsigned char *newVariable = (char* ) passedVariable;

这是为什么正在做？我总是尽量在使用符号/无符号一致的，因为我知道，这两者之间的切换可能会造成问题，但是这似乎开发商并没有在意。

Why is this being done? I always try to be consistent in the use of signed/unsigned, because I know that switching between the two can cause problems, but this developer doesn't seem to care.

推荐答案

更改指针类型是不是一个真正的问题，该地址仍然有效。
然而除preting尖头数据符号/无符号使得当且仅当...签名的数据为负值的差值。因此，在你的例子，如果你的字符的始终是肯定的，那么它的确定，否则就不是。

Changing the pointer type is not really an issue, this address will still be valid. However interpreting the pointed data as signed/unsigned makes a difference if and only if... the signed data is negative. So in your example if your char's are always positive, then it's ok, otherwise it is not.

符号/无符号的类型转换的例子：

Example of signed/unsigned casts:

char c = 42;
char d = -42;
unsigned char cu = c;
unsigned char du = d;

printf("c  %d\n", c);
printf("cu %d\n", cu);
printf("d  %d\n", d);
printf("du %d\n", du);

输出：

c  42
cu 42
d  -42
du 214

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