为什么一个字符在数组中似乎比它本身占用更多的空间?

char testChar = 'a';
char myCharString[] = "asd";
char *pointerToFirstChar = &(myCharString[0]);
char *pointerToSecondChar = &(myCharString[1]);
cout << "A char takes " << sizeof(testChar) << " byte(s)";
cout << "Value was " << pointerToFirstChar << ", address: " << &pointerToFirstChar << endl;
cout << "Value 2 was " << pointerToSecondChar << ", address:" << &pointerToSecondChar << endl;

this outputs:

这个输出:

"A char takes 1 byte"

" char取1个字节"

"... address: 00F3F718"

“…地址:00 f3f718”

"... address: 00F3F70C",

“…地址:00 f3f70c”,

I'm thinking the difference between addresses should be 1 byte, since that would be the size of the data separating them. Why is it not so?

我认为地址之间的区别应该是1个字节，因为这是数据分离的大小。为什么不是这样呢?

3 个解决方案

#1

&pointerToFirstChar and &pointerToSecondChar, you're not taking the address of the elements of the char array, but the address of the local variable pointerToFirstChar and pointerToSecondChar. Note they've been pointers themselves.

&pointerToFirstChar和&pointerToSecondChar，您没有获取char数组元素的地址，而是本地变量pointerToFirstChar和pointerToSecondChar的地址。注意，它们都是指针。

You might want:

你可能会想:

cout << "Value was " << pointerToFirstChar << ", address: " << static_cast<void*>(pointerToFirstChar) << endl;
cout << "Value 2 was " << pointerToSecondChar << ", address:" << static_cast<void*>(pointerToSecondChar) << endl;

Note you need to cast them to void* to print out the address instead of the string.

注意，您需要将它们转换为void*以打印地址而不是字符串。

#2

You're looking at the addresses of the pointers pointerToFirstChar and pointerToSecondChar. They are pointers to char; compare their values and those will differ by 1. You seem to have edited that out in your text.

您正在查看指针pointerToFirstChar和pointerToSecondChar的地址。它们是指向char的指针;比较它们的值，它们的值会相差1。你似乎已经在你的文章中编辑过了。

#3

you are printing address of pointer variable and not the address currently pointer is holding.

你打印的是指针变量的地址，而不是当前指针所在的地址。

for example:

例如:

&myCharString[0] = 0xFE20  
&myCharString[1] = 0xFE21  
&myCharString[2] = 0xFE23

char *pointerToFirstChar = &(myCharString[0]);

Address of pointerToFirstChar = 0xF8C2 and it holds the address &myCharString[0] = 0xFE20

pointerToFirstChar = 0xF8C2的地址，它保存地址&myCharString[0] = 0xFE20。

so you are printing, 0xF8C2 rather than printing 0xFE20

所以你要打印，0xF8C2而不是打印0xFE20。

Update your code as follows to get the correct result.

按照以下方法更新代码以获得正确的结果。

cout << "Value was " << pointerToFirstChar << ", address: " << (void *)&pointerToFirstChar[0] << endl; cout << "Value 2 was " << pointerToSecondChar << ", address:" << (void *)&pointerToSecondChar[0] << endl;

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For more information please follow the below link print address of array of char

有关更多信息，请遵循以下链接打印地址数组的char。

#1