如何通过内存分配特性来度量int和short变量的大小?

I tried to test the length of int and short type in my computer (X86_64), so I wrote two pieces of codes as bellow: short:

我尝试在我的计算机(X86_64)中测试int和short类型的长度，所以我写了两段代码:short:

short a;
scanf("%hd%hd",&a,&a+1);
printf("%hd",a+*(&a+1));

int:

int(警官):

int a;
scanf("%d%d",&a,&a+1);
printf("%d",a+*(&a+1));

Both work and output the right answer, but when I change to number 1 to 4, the first one works well while the second one show a Segmentation fault after input.
Some materials say that the lengths of short and int are all 16 bits in x86, I don't know the difference in x86_64, are they the same? In addition, what caused the Segmentation fault?

工作和输出都是正确的答案，但是当我改变到数字1到4时，第一个工作很好，第二个在输入后显示一个分割错误。有些材料说，在x86中，短和整数的长度都是16位，我不知道x86_64的区别，它们是一样的吗?此外，是什么导致了分割错误?

PS:I use gcc -Wall -O2 -o filename -lm to compile.

我使用gcc -Wall -O2 -o文件名-lm来编译。

3 个解决方案

#1

All x86 ABIs define sizeof(short) == 2.
All x86 16-Bit ABIs define sizeof(int) == 2, all other x86 ABIs define sizeof(int) == 4. For standard 64Bit mode, every pointer is 64Bit == 8 Byte big, for 64Bit 4G-Address mode and for 32Bit flat memory mode (flat is standard), every pointer is 32Bit == 4 Byte, and for 16Bit mode it varies.

所有x86 ABIs定义sizeof(short) == 2。所有x86 16位ABIs定义sizeof(int) == 2，所有其他x86 ABIs定义sizeof(int) == 4。对于标准的64位模式，每个指针都是64位== 8字节的大，对于64位的4g地址模式和32位平的内存模式(单位是标准的)，每个指针都是32位== 4字节，而对于16位模式则是不同的。

This example has Undefined Behavior, because you do not own a short-object at &a+1:

这个例子有未定义的行为，因为你没有一个短对象在和+1:

short a;
scanf("%hd%hd",&a,&a+1);
printf("%hd",a+*(&a+1));

Same for this one, just with int:

对于这个也是一样，只是用int:

int a;
scanf("%d%d",&a,&a+1);
printf("%d",a+*(&a+1));

Undefined Behavior means anything may happen, even the proverbial nasal demons.
That includes it seems to work as well as reformatted your drive.

未定义的行为意味着任何可能发生的事情，甚至是众所周知的鼻魔。这包括它似乎工作和重新格式化你的驱动器。

#2

You can get the size of short by:

你可以通过以下方式获得:

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

You can get the size of short pointer, by:

你可以得到短指针的大小:

printf("size of pointr: %d\n", sizeof(short*));

I believe your other questions have been answered by Deduplicator.

我相信你的其他问题已经被de复印机解决了。

#3

I have known what's wrong with my code after reading your comments and answers.
I guessed the address of a variable is like that in assembly language, so if I get an address and add it with the length of a variable I can get the address of the following viable. I use &a+4 first to test if the length of int is 4 BYTES, but there was a Segmentation fault while running, I changed 4 to 1 and get the correct value, I didn't understand why this worked? Now, I got it.
In the following code:

在阅读了您的评论和答案后，我已经知道我的代码有什么问题。我猜一个变量的地址就像汇编语言中那样，所以如果我得到一个地址并加上一个变量的长度，我就能得到下面可行的地址。我先用and +4来测试int的长度是否为4个字节，但是在运行的时候有一个分割错误，我将4改为1并得到正确的值，我不明白为什么会这样?现在,我明白了。在以下代码:

short a,b;
a = b = 1024;
cout << &a << endl << &b << endl;
cout << &a << endl << (&a) + 1;

An output of it is:

它的输出是:

0xbfad153c  
0xbfad153e  
0xbfad153c  
0xbfad153e

And

和

int a,b;
a = b = 1024;
cout << &a << endl << &b << endl;
cout << &a << endl << (&a) + 1;
return 0;

An Output is:

一个输出是:

0xbfddb968  
0xbfddb96c  
0xbfddb968  
0xbfddb96c

It shows that &a + 1 here plus the value of the size of int a but not the literal const value of 1
So I just want to use the feature of memory allocating to measure the size of int and short, not sizeof function.
Thanks for your answers and comments, even though they are not what I want to get.

它显示了这里的&a + 1加上int a大小的值，而不是常量1的值，所以我想用内存分配的特性来测量int和short的大小，而不是sizeof函数。谢谢你的回答和评论，尽管它们不是我想要的。

#1