C / C ++：在32位计算机上与在64位计算机上相比，sizeof（short），sizeof（int），sizeof（long），sizeof（long long）等... [英] C/C++: sizeof(short), sizeof(int), sizeof(long), sizeof(long long), etc... on a 32-bit machine versus on a 64-bit machine

问题描述

我正在运行Windows 7（64位）。

此问题与在此找到的问题相同：

在64位计算机上很长

，但是它更深入，因为它处理甚至更多的数据类型，并将
应用于C或C ++，而不是C＃。首先，我正在使用Microsoft Visual Studio Ultimate2012。
不幸的是，尽管此IDE支持C＃和Visual C ++，但它似乎不再支持普通的
旧VisualC。无论如何，我尝试在IDE中创建以下标准C ++
程序：

  #include< cstdio> ; 
 
 int main（int argc，char ** argv）{
 
 printf（ sizeof（short）：％d\n，（int）sizeof（short） ）; 
 
 printf（ sizeof（int）：％d\n，（int）sizeof（int））; 
 
 printf（ sizeof（long）：％d\n，（int）sizeof（long））; 
 
 printf（ sizeof（long long）：％d\n，（int）sizeof（long long））; 
 
 printf（ sizeof（size_t）：％d\n，（int）sizeof（size_t））; 
 
 printf（ sizeof（void *）：％d\n，（int）sizeof（void *））; 
 
 printf（ Hit enter to exit.\n）; 
 
 char * scannedText; 
 
 scanf（％s，& scannedText）; 
 
返回0; 
 
} 
  

，由于找不到运行a的选项控制台应用程序，我只是
在返回0处设置了一个断点；语句，以便在控制台中查看输出
。结果是：

  sizeof（short）：％d\n，4 
 sizeof（int）： ％d\n，4 
 sizeof（long）：％d\n，4 
 sizeof（long long）：8 
 sizeof（size_t）：4 
 sizeof（void *）：4 
按Enter退出。
  

旧C教科书指出int设置为字长，在16位
计算机上为16，在32位计算机上为32，但是在64位
系统上，该规则似乎中断了字长为64。相反，我从
那里读到的这些系统就像32位系统，但是对
64位计算的支持要比其32位系统好。 ，从上述C ++程序获得的结果
与在32位系统上获得的结果
完全相同。数据类型（size_t）的大小（可以将
用于测量内存中对象占用的内存量）
还将其值存储在4个字节中，并且有趣的是
的大小用于访问备忘录ry位置（例如sizeof（void *）
显示用于将通用指针存储到内存中任何位置的位数）
也是32位长。

任何人都知道Visaul C是如何从Visual Studio 2012中删除的，以及是否
是否仍然可以从Visual Studio 2012运行控制台应用程序而无需
设置断点或从标准中读取文本

此外，我的解释是正确的，还是我配置错误？在IDE中使用b $ b，例如，它是为32位而不是64位
系统编译的？根据发布者之一的说法，由于我的系统是64位的，因此我应该
看到此处针对size_t和指针描述的结果：
https://en.wikipedia.org/wiki/64-bit_computing#64-bit_data_models
但我没有看到。有没有一种方法可以重新配置Visual Studio
，使其可以支持64位内存模型，而不是我在程序输出中当前看到的
？ b $ b

谢谢。

解决方案

对我来说很对。在c / c ++中，int并不是根据位大小来专门定义的。创建项目时，可以选择控制台应用程序。 VS2012仍然支持C，但是它们大多将项目集成到C / C ++中。有一个编译器选项（我认为是/ TC），它将强制编译器符合C规范。默认情况下，它将通过文件扩展名隐含该语言。 MS C支持不是理想的，例如它不包括stdbool.h。

如果要控制数据的位大小，可以使用stdint。 h，其中包含确切的宽度int数据类型。

I'm running Windows 7 (64-bit).

This question looks at the same question found here:

long on a 64 bit machine

but is more in-depth as it deals with even more data types and applies to C or C++, not C#. First of all, I am using Microsoft Visual Studio Ultimate 2012. Unfortunately, while this IDE supports C# and Visual C++ it no longer supports plain old Visual C it seems. Anyhow, I've tried the creating the following standard C++ program in the IDE:

#include <cstdio>

int main(int argc, char **argv) {

  printf("sizeof(short): %d\n", (int) sizeof(short));

  printf("sizeof(int): %d\n", (int) sizeof(int));

  printf("sizeof(long): %d\n", (int) sizeof(long));

  printf("sizeof(long long): %d\n", (int) sizeof(long long));

  printf("sizeof(size_t): %d\n", (int) sizeof(size_t));

  printf("sizeof(void *): %d\n", (int) sizeof(void *));

  printf("Hit enter to exit.\n");

  char *scannedText;

  scanf("%s", &scannedText);

  return 0;

}

and since I couldn't find the option to run a console application I simply placed a breakpoint at the "return 0;" statement, so as to view the output in the console. The result was:

sizeof(short): %d\n", 4
sizeof(int): %d\n", 4
sizeof(long): %d\n", 4
sizeof(long long): 8
sizeof(size_t): 4
sizeof(void *): 4
Hit enter to exit.

Old C textbooks state that int is set to the "word size", which is 16 on 16-bit machines and 32 on 32-bit machines. However this rule seems to break on 64-bit systems where one would expect the "word size" to be 64. Instead, from what I've read these systems are like 32-bit systems but have better support for 64-bit computations than their 32-bit counterparts did. Hence, the results obtained from the above C++ program are exactly the same as one would obtain on a 32-bit system. The size of data types (size_t) (which can be used to measure amount of memory taken up by objects in memory) also stores its values in 4 bytes, and it is also interesting that the size of pointers used to access memory locations (for instance sizeof(void *) shows the number of bits used to store generic pointers to any location in memory) is also 32 bits long.

Anyone know how come Visaul C was removed from Visual Studio 2012 and whether it is still possible to run console applications from Visual Studio 2012 without having to set a breakpoint or read text from standard input prior to exiting as above in order for the console window to pause before closing?

Furthermore, is my interpretation correct, or do I have something misconfigured in the IDE so that, for instance, it compiles for 32-bit rather than for 64-bit systems? According to one of the poster, since my system is 64-bit, I should see the results described here for size_t and pointers: https://en.wikipedia.org/wiki/64-bit_computing#64-bit_data_models but I am not seeing this. Is there a way to reconfigure Visual Studio so that it may support a 64-bit memory model, as opposed to what I am currently seeing in the program's output?

Thanks.

解决方案

Looks right to me. In c/c++ int isn't specifically defined in terms of bit-size. When creating a project you can select a "console application". VS2012 still supports C, but they mostly lump projects into C/C++. There is a compiler option (/TC I think) which will force the compiler into C compliance. By default it will imply the language by the file extension. MS C support isn't ideal, it doesn't include stdbool.h for instance.

If you want to control the bit size of your data you can use stdint.h which contains exact width int datatypes.

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