当左操作数未定义行为为负 [英] undefined behavior when left operand is negative

问题描述

几天就回来我给微软GD在线考试实习那里。我一直研究了负数的左移是不确定的行为，但该文件几乎7个问题出30相关的转向运营商和大约5个问题分别疗法这涉及到转向负数离开，他们别无选择，说未定义行为。我很震惊地看到这一点。所以，我的问题是，有这个C标准改变了吗？这是目前定义的？
样本问题：

few days back I gave Microsoft GD online exam for internship there. I have always studied that the left shift of negative number is an undefined behavior but that paper had almost 7 questions out of 30 related to shift operators and approx 5 questions were ther which involved shifting negative numbers to left and they had no option saying "undefined behavior". I was shocked to see that . So, my question is that has this C standard changed ? Is this defined now ? sample question :

printf("%d",-1<<10);

I marked its answer as -1024 by the logic 2^10*-1

我甚至跑这对gcc和它给了我O / P为-1024 NLY（当我回到了家。）

I even ran this on gcc and it gave me o/p as -1024 nly (when I returned home.)

推荐答案

规则并没有改变。它仍然在技术上不确定的。

The rules haven't changed. It's still technically undefined.

从C标准引用（第6.5.7条第4款n1548的）：

Quoting from the C standard (Section 6.5.7, paragraph 4, of n1548):

E1 1所述的结果;＆下; E2是E1左移E2位位置;腾空位音响与LLED
  零。如果E1有一个无符号类型，则结果的值是E1×2 ^ E2，减小模数比在结果类型的最大值重新presentable一个。如果E1有符号类型和非负价值，在E1结果类型×2 ^ E2重新presentable，然后就是
  所得到的值; ，否则，其行为定义理解过程网络。

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

它清楚地说，如果 E1 不签名或与非负值签署，该行为是不确定的。

It clearly says if E1 is not unsigned or signed with nonnegative value, the behavior is undefined.

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