Unexpected results for “sizeof” of strings

问题

Why would sizeof in the following cases print different values:

printf("%d",sizeof("ab")); //print 3

char* t="ab";
printf("%d",sizeof(t)); //print 4

In the first case I have 2 characters... Shouldn't sizeof print 2? Because they are 2 bytes?

回答1:

Strings in C are null terminated.

"ab" in memory looks like 'a' 'b' '\0'

While t is a pointer, so size is 4.

回答2:

t is a pointer to an array containing "ab". Its size is the size of a pointer.

"ab" is an array containing "ab". Its size is the size of that array, which is three characters because you have to account for the null terminator.

Arrays are not pointers.

回答3:

Because in the first case you are asking for the sizeof an array. The second time you are asking for the size of a pointer.

回答4:

A string literal is a char array, not a char *.

char a[] = "ab";
char * t = a;
printf("%d",sizeof(a)); //print 3
printf("%d",sizeof(t)); //print 4

回答5:

The type of the string literal "ab" is const char(&)[3].
So sizeof("ab") = sizeof(const char(&)[3]) = sizeof(char) * 3 which is 3 on your machine.

And in the other case, t is just a pointer.
So sizeof(t) = sizeof(void*) which is 4 bytes on your machine.

Note:

If you prepend "ab" with L, and make it L"ab", then,

The type of the string literal L"ab" is const wchar_t(&)[3].
So sizeof(L"ab") = sizeof(const wchar_t(&)[3]) = sizeof(wchar_t) * 3 which is 12 on ideone:

http://ideone.com/IT7aR

So that is because sizeof(wchar_t) = 4 on ideone which is using GCC to compile!

来源：https://stackoverflow.com/questions/5463325/unexpected-results-for-sizeof-of-strings

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c++

sizeof