不了解基本的C问题 [英] Not understanding basic C problem

问题描述

我期待答案应该是1.但是在GCC编译器中，Turbo c编译器将结果显示为0.在某些DEV c编译器中显示结果为1.哪一个是正确的，为什么？

I am expecting the answer should come as 1.But in GCC compiler and Turbo c compiler is showing the result as 0.In some DEV c compiler showing the result as 1.Which one is correct and why ?

#include<stdio.h>
int main()
{
   int z = 0;
   z = 0 || z++;
   printf ("%d", z);
   return 0;
 }

我尝试过：



我在Dev c，GCC和Turbo c编译器中尝试过。

What I have tried:

I have tried in Dev c , GCC and Turbo c compiler.

推荐答案

你有一个逻辑短路评估 - 维基百科 [ ^ ]表单的操作

You have a logical Short-circuit evaluation - Wikipedia[^] operation of the form

a || b

第一个表达式的布尔结果

The boolean result of the first expression

z = 0

是赋值后左操作数 z 的值。当它为零时，布尔结果为false。你可以检查这个：

is the value of the left operand z after the assignment. When that is zero, the boolean result is false. You can check this:

int z = 0;
int a = (z = 0) ? 1 : 0;
int b = (z = 1) ? 1 : 0;
printf("%d %d\n", a, b);

这应打印 0 1 包含所有编译器。



一旦评估了左项的布尔结果，第二次操作只应在第一次操作时执行：

This should print 0 1 with all compilers.

Once the boolean result of the left term has been evaluated, the second operation should be only executed when the first was true:

int b = (z = 0) ? 1 : 0;
if (b)
    b = (z++) ? 1: 0;

然后 z 将仍然为零，因为未执行增量操作。



以上是错误的（谢谢NV）！

如果，它应该是偏离的！b）

[/ EDIT2]



因此执行第二次操作的编译器可能是不符合C标准。我写了也许因为我不确定在这种特殊情况下没有使用逻辑结果。



您可以这样检查（所有优化都被禁用） ;例如-O0 with GCC）：

Then z will be still zero because the increment operation is not executed.

The above is wrong (thank you NV)!
It should be off course if (!b)
[/EDIT2]

So compilers executing always the second operation may be not conforming to the C standard. I wrote "maybe" because I'm not really sure in this special case where the logical result is not used.

You might check it this way (all optimisations disabled; e.g. -O0 with GCC):

if (z = 0 || z++)
    printf("true: %d\n", z); // Never reached
else
    printf("false: %d\n", z); // Always executed

编译器这里打印1不符合C标准。



另请注意，优化编译器可能会检测到逻辑操作的结果始终为false并将零分配给 z 甚至跳过该行

A compiler that prints "1" here is not conforming to the C standard.

Note also that optimising compilers might detect that the result of the logical operation is always false and assign zero to z or even skip the line

z = 0 || z++;

因为 z 已经在上一行中设置为零。

[/ EDIT]





解决方案3指出了问题。

行为未定义，如警告所示。

试试这个：

because z has been already set to zero in the previous line.
[/EDIT]


Solution 3 is pointing out the problem.
The behaviour is undefined as indicated by the warning.
Try this:

int z = 0;
z = 0 || z++;
printf ("z: %d\n", z);

(z = 0) || (z++);
printf ("z: %d\n", z);

使用GCC，它将打印

With GCC, it will print

z: 0
z: 1

[/ EDIT2]

[/EDIT2]

gcc -Wall mytest.c
mytest.c: In function ‘main’:
mytest.c:6:7: warning: operation on ‘z’ may be undefined [-Wsequence-point]
     z = 0 || z++;

那就是你遇到 C 编程语言的一个黑暗角落。例如，请参见序列点 - 维基百科 [ ^ ]。

作为附注，启用所有警告（ -Wall gcc 可能证明它本身很有用。

That is you hit a dark corner of the C programming language. See, for instance Sequence point - Wikipedia[^].
As a side note, enabling all the warnings (-Wall with gcc may prove itself useful, after all).

你对这些陈述有什么期望：

What did you expect from the statements:

int z = 0;
z = 0 || z++;

这又是在z分配之前或之后是否执行后增量的问题。

This is again a question of whether the post-increment is performed before or after the assignment to z.

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