在数组中打印元素[复制]

Possible Duplicate:
Confused about C macro expansion and integer arithmetic
A riddle (in C)

可能重复:对C宏展开和整数算法感到困惑

The output of the following C program is to print the elements in the array. But when actually run, it doesn't do so.

下面C程序的输出是打印数组中的元素。但实际运行时，它不会这么做。

  #include<stdio.h>

  #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
  int array[] = {23,34,12,17,204,99,16};

  int main()
  {
      int d;

      for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
          printf("%d\n",array[d+1]);

      return 0;
  }

Why is that?

这是为什么呢?

7 个解决方案

#1

This fails because sizeof is returning a value of type size_t, which is unsigned. This causes the comparison to promote the -1 to unsigned, which is generally a very large value and thus make the comparison fail.

这之所以失败，是因为sizeof返回的是一个类型size_t的值，它是无符号的。这导致比较将-1提升为无符号，这通常是一个很大的值，从而使比较失败。

You should receive warnings for the sign mismatch.

您应该收到符号不匹配的警告。

#2

sizeof() returns unsigned int and so in the <= comparison, d gets converted to unsigned int. -1 when converted to unsigned int is 0xffffff. Since 0xffffff > (7-2), for-loop body won't be executed even once.

sizeof()返回无符号整型，因此在<=比较中，d被转换为无符号整型。由于0xffffff > (7-2)， for循环体甚至不会执行一次。

#3

The point is the type of value sizeof returns.

关键是返回值sizeof的类型。

It's a size_t, which is an unsigned type, so converting -1 to that type for the comparison produces the maximal value of the type, which is far larger than 5.

它是一个size_t，它是一个未签名的类型，所以将-1转换为该类型的比较会产生类型的最大值，这比5大得多。

#4

for(d=-1;d <= ((int)TOTAL_ELEMENTS-2);d++)

#5

I guess this is more clear:

我想这更清楚了:

for(d = 0; d < TOTAL_ELEMENTS; d++)
   printf("%d\n",array[d]);

And size_t case should not be a problem anymore.

size_t应该不再是个问题了。

#6

Your original problem works, when I do the following:

当我做以下工作时，你的原始问题是有效的:

#define TOTAL_ELEMENTS 7
int array[] = {23,34,12,17,204,99,16};

int main()
{
  int d;

  for(d=-1; d <= (TOTAL_ELEMENTS - 2); d++)
      printf("%d\n",array[d+1]);

  return 0;
}

This has something to do with comparing a negative number d=-1 with (TOTAL_ELEMENTS-2). I believe it might be sizeof, butI will leave finding out why as an exercise.

这与将负数d=-1与(TOTAL_ELEMENTS-2)进行比较有关。我相信它可能是尺寸的，但我将把它作为一种练习。

Leaving your original #define as it was --

保留原来的#define

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))

#定义TOTAL_ELEMENTS (sizeof(数组)/ sizeof(数组[0]))

-- but using zero-based indexing also worked.

——但是使用基于零的索引也是有效的。

I'm not sure why your original TOTAL_ELEMENTS definition caused the loop not to print.

我不确定为什么原始的TOTAL_ELEMENTS定义导致循环不打印。

#7

A more standard way to write this loop which works would be

一个更标准的方法来写这个循环，它是可行的

for(d = 0; d < TOTAL_ELEMENTS; d++)
   ...

as was pointed out by correctly by several others here already, the problem lies with signed/unsigned values being compared. This construct would have avoided the problem.

正如这里的其他几个人已经正确地指出的那样，问题在于比较有符号/无符号的值。这个构造可以避免这个问题。

#1