问题描述
如何检测 INT 签署岬用C？
这问题大多是历史的机器。我所问如何，如果一个整数为0或-0区别。在1的补数和符号/幅度INT编码，既是一个0（或+0）和-0是可能的。
简单的符号位的测试是比较反对 0 。
INT X;
的printf（符号位为％s \\ n，（X℃下）设置：未设置）;
但是，这无法在1的补，并签署幅度时 X 是 -0 。
1号候选方法：模板测试结果
由于C定义，一个 INT 必须有一个符号位，无论整数编码，下面应该工作。
INT X;
INT SignBitMask =待定;
的printf（符号位为％s \\ n，（X安培; SignBitMask）设置：未设置）;
问题就变成如何确定 SignBitMask 的C中的价值？结果 SignBitMask = INT_MAX + 1 似乎是一个起点。
第二候选方法：创建功能和校验位模式：
INT IsSignBitSet（INT X）{
如果（X 0）返回0;
如果（X℃，）返回1;
INT ZP = 0;
如果（memcmp（安培; X，＆安培; ZP，sizeof的X）== 0）返回0;
INT锌= -0; //这甚至形成-0的方式吗？
如果（memcmp（安培; X，＆安培;锌，sizeof的X）== 0）返回1;
//如果我们来到这里，现在该怎么办？
返回？;
}
我想没有便携统一的解决办法 - 也许因为需要不再存在
为什么：我想知道如何把不同进行检测，并签署了印制零
请注意：我特意在这里避免了C标记，并想我第一次尝试就在历史记录标签
。答
组合的3答案和C11dr 6.2.6.2整型（信息为 INT ，一个符号位的必须的存在，积极的符号位为0时，负符号位为1），溶液（出现独立1的补，2的补码和符号/幅值整数编码）是
INT IsSignBitSet_Best（INT X）{
//返回1，如果x小于0 _or_ x是算术0与一些位集。
返回（X小于0）|| （（X == 0）及及（*（（unsigned int类型*）及X）））;
}
直接面具的方法是最简单的，但还没有拿出一个高度便携面具定义
INT IsSignBitSet_Simple（INT X）{
静态无符号SignBitMask = 0x80的; //或者其他一些平台依赖面膜
返回（（符号）x＆安培; SignBitMask）！= 0;
}
要寻找负0对零只是检查与所有设置任何位。
INT testForNegative0（INT X）{
则返回（x == 0安培;＆放大器; *（（unsigned int类型*）及X））;
}
或回答标题中的问题：
INT hasSignBitSet（INT X）{
返回（X小于0）|| testForNegative0（X）;
}
这适用于你提到的3编码，它可能不适合更深奥的那些工作。
How to detect an int sign-ness in C?
This question is mostly of historical machines. What I am asking how to distinguish if an integer is 0 or -0. In 1's complement and sign/magnitude int encoding, both a 0 (or +0) and -0 are possible.
The simple sign bit test is to compare against 0.
int x;
printf("sign bit is %s\n", (x < 0) ? "set" : "not set");
But this fails in 1's complement and sign magnitude when x is -0.
1st Candidate approach: Mask test.
As C defines that an int must have a sign bit regardless of integer encoding, the following should work.
int x;
int SignBitMask = tbd;
printf("sign bit is %s\n", (x & SignBitMask) ? "set" : "not set");
The question becomes how to determine the value of SignBitMask in C?SignBitMask = INT_MAX + 1 seems like a starting point.
2nd Candidate approach: create function and check bit patterns:
int IsSignBitSet(int x) {
if (x > 0) return 0;
if (x < 0) return 1;
int zp = 0;
if (memcmp(&x, &zp, sizeof x) == 0) return 0;
int zn = -0; // Is this even the way to form a -0?
if (memcmp(&x, &zn, sizeof x) == 0) return 1;
// If we get here, now what?
return ?;
}
I'm thinking there is no portable uniform solution - maybe because the need no longer exists.
Why: I have wondered how various signed zeros were detected and printed.
Note: I have purposely avoided the "C" tag here and thought I'd try just the "History" tag first.
[Edit] Answer
Combining info of 3 answers and C11dr 6.2.6.2 "Integer types" (for int, a single sign bit must exist, the positive sign bit is 0, the negative sign bit is 1), a solution (that appears independent of 1's complement, 2's complement and sign/magnitude integer encoding) is
int IsSignBitSet_Best(int x) {
// return 1 if x is less than 0 _or_ x is arithmetically 0 with some bit set.
return (x < 0) || ((x == 0) && (* ((unsigned int*) &x) ));
}
The direct mask approach is simplest, but have not come up with a highly portable mask definition
int IsSignBitSet_Simple(int x) {
static unsigned SignBitMask = 0x80; // Or some other platform dependent mask
return ((unsigned)x & SignBitMask) != 0;
}
To find negative 0's just check for a zero with any bit at all set.
int testForNegative0(int x) {
return (x==0 && *((unsigned int*)&x));
}
Or to answer the question in the title:
int hasSignBitSet(int x) {
return (x<0) || testForNegative0(x);
}
This works for the 3 encodings you mention, it may not work for even more esoteric ones.
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