为什么通过使用printf(％x)来使指针的值不同? [英] Why the value of pointer is different by using printf(%x)?

问题描述

#include <stdio.h>
int main(void)
{
  int *ptr;
  printf("The Hex value of ptr is 0x%x",ptr);
  printf("The pointer value of ptr is %p",ptr);
}

与输出有点不同，我不知道为什么

and the output is a little different that I don't know why

The Hex value of ptr is 0x24a77950
The pointer value of ptr is 0x7fff24a77950

它显示ptr的值是一个十六进制整数，但十六进制输出缺少部分7fff.

It shows the value of ptr is a hex integer, but the hex output lack the part 7fff.

这是printf格式问题还是其他原因?

Is this the printf formatting issue or something else?

推荐答案

％x将指针转换为无符号整数(32位长度).在64位计算机上，指针的长度为8字节(64位).

%x casts your pointer to an unsigned integer (32-bit length). On a 64-bit machine your pointer is of 8-byte (64 bit) length.

使用％p进行打印将以其完整大小-64位打印整个指针.但是，当您使用％x进行打印时，仅低32位被打印.因此，使用％p打印指针始终是安全的.

Printing with %p prints the whole pointer, in its complete size - 64 bits. But when you are printing with %x, only the lower 32 bits are printed. Hence it is always safe to print a pointer with %p.

您可以添加以下两行并进行验证:

You can do add extra 2 lines as below and verify:

printf("size of unsigned int is %lu\n", sizeof(unsigned int));
printf("size of pointer is %lu\n", sizeof(int *));

在具有64位操作系统的64位计算机上，这应该分别给您4和8.

On a 64-bit machine with a 64-bit operating system, this should give you 4 and 8 respectively.

查看已回答的两个类似问题:问题1 和<一个href ="https://stackoverflow.com/questions/399003/is-the-sizeofsome-pointer-always-equal-to-four">问题2

See two similar questions that have been answered: question 1 and question 2

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