问题
printf("%lu \n", sizeof(*"327"));
I always thought that size of a pointer was 8 bytes on a 64 bit system but this call keeps returning 1. Can someone provide an explanation?
回答1:
Putting * before a string literal will dereference the literal (as string literal are array of characters and will decay to pointer to its first element in this context). The statement
printf("%zu \n", sizeof(*"327"));
is equivalent to
printf("%zu \n", sizeof("327"[0]));
"327"[0] will give the first element of the string literal "327", which is character '3'. Type of "327", after decay, is of char * and after dereferencing it will give a value of type char and ultimately sizeof(char) is 1.
回答2:
The statement:
printf("%lu \n", sizeof(*"327"));
actually prints the size of a char, as using * dereferences the first character of string 327. Change it to:
char* str = "327";
printf("%zu \n", sizeof(str));
Note that we need to use %zu here, instead of %lu, because we are printing a size_t value.
回答3:
The string literal is an anonymous, static array of chars, which decays to a pointer to its first character -- that is, a pointer value of type char *.
As a result expression like *"abc" is equivalent to *someArrayOfCharName, which in turn is equivalent to *&firstCharInArray which results in firstCharInArray. And sizeof(firstCharInArray) is sizeof(char) which is 1.
回答4:
Good answer by haccks.
Also, the behaviour of your code is undefined, because you have used the wrong format specifier.
So, use %zu instead of %lu because sizeof() returns size_t and size_t is unsigned.
C11 Standard: §7.21.6.1: Paragraph 9:
If a conversion specification is invalid, the behavior is undefined.225) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.
来源：https://stackoverflow.com/questions/46705264/why-does-sizeof327-return-1-instead-of-8-on-a-64-bit-system