是< a> = b"相当于“a - b> = 0”？ [英] Is "a >= b" equivalent to "a - b >= 0"?
问题描述
嗨！
表达式是a> = b相当于a - b> = 0在C / C ++中？
这个等价是IEEE / ANSI规则吗？或者这台机器/编译器
是否依赖？
欢迎任何参考！
先谢谢，Humberto 。
Hi!
Is the expression "a >= b" equivalent to "a - b >= 0" in C/C++?
Is this equivalence an IEEE/ANSI rule? Or is this machine/compiler
dependent?
Any references are welcome!
Thanks in advance, Humberto.
推荐答案
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嗨！
表达式是a> = b相当于a - b> = 0在C / C ++中？
这个等价是IEEE / ANSI规则吗？或者这台机器/编译器是否依赖？
欢迎任何参考文献！
Hi!
Is the expression "a >= b" equivalent to "a - b >= 0" in C/C++?
Is this equivalence an IEEE/ANSI rule? Or is this machine/compiler
dependent?
Any references are welcome!
在C中，考虑如果发生以下情况： b == MIN_INT。
在C ++中，考虑任何运算符都可能过载，并且在每种情况下调用
不同的[set of]运算符。
- Risto -
In C, consider what happens if e.g. b==MIN_INT .
In C++, consider that any operator could be overloaded, and that
different [set of] operators is called in each case.
- Risto -
不一样。
考虑（ ab）小于可能的最小值。
特殊情况是无符号类型。
marc
It is not the same.
Consider (a-b) is smaller than the smallest possible value.
A special case of this are unsigned types.
marc
< hj ****** @ mat.puc-rio.br>：
<hj******@mat.puc-rio.br>:
表达式是a> = b ;相当于a - b> = 0在C / C ++中？
这个等价是IEEE / ANSI规则吗？或者这台机器/编译器是否依赖？
Is the expression "a >= b" equivalent to "a - b >= 0" in C/C++?
Is this equivalence an IEEE/ANSI rule? Or is this machine/compiler
dependent?
如果a和b未签名，那么当然：不！
#include< stdio.h>
int main（）
{
unsigned a = 1，b = 2;
printf（"％d％c =％d \ n"，> = b，"！=" [（a> = b）==（a - b> ; = 0）]，a - b> =
0）;
}
即使a和b也是如此是整数，答案是：不！
#include< stdio.h>
#include< limits.h>
int main（）
{
int a = INT_MIN，b = INT_MAX;
printf（"％d％c = ％d \ n"，a> = b，"！=" [（a> = b）==（a - b> = 0）]，a - b> =
0）;
}
Jirka
If a and b are unsigned, then of course : No!
#include <stdio.h>
int main()
{
unsigned a = 1, b = 2;
printf("%d %c= %d\n", a >= b, "!="[(a >= b) == (a - b >= 0)], a - b >=
0);
}
and even if a and b are ints, the answer is: No!
#include <stdio.h>
#include <limits.h>
int main()
{
int a = INT_MIN, b = INT_MAX;
printf("%d %c= %d\n", a >= b, "!="[(a >= b) == (a - b >= 0)], a - b >=
0);
}
Jirka
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