问题
Why is size of this char variable equal 1?
int main(){
char s1[] = "hello";
fprintf(stderr, "(*s1) : %i\n", sizeof(*s1) ) // prints out 1
}
回答1:
NOTA: the original question has changed a little bit at first it was: why is the size of this char pointer 1
sizeof(*s1)
is the same as
sizeof(s1[0]) which is the size of a char object and not the size of a char pointer.
The size of an object of type char is always 1 in C.
To get the size of the char pointer use this expression: sizeof (&s1[0])
回答2:
Why is size of this char variable equal 1?
Because size of a char is guaranteed to be 1 byte by the C standard.
*s1 == *(s1+0) == s1[0] == char
If you want to get size of a character pointer, you need to pass a character pointer to sizeof:
sizeof(&s1[0]);
回答3:
Because you are deferencing the pointer decayed from the array s1 so you obtain the value of the first pointed element, which is a char and sizeof(char) == 1.
回答4:
sizeof(*s1) means "the size of the element pointed to by s1". Now s1 is an array of chars, and when treated as a pointer (it "decays into" a pointer), dereferencing it results in a value of type char.
And, sizeof(char) is always one. The C standard requires it to be so.
If you want the size of the whole array, use sizeof(s1) instead.
回答5:
sizeof(*s1) means its denotes the size of data types which used. In C there are 1 byte used by character data type that means sizeof(*s1) it directly noticing to the character which consumed only 1 byte.
If there are any other data type used then the **sizeof(*data type)** will be changed according to type.
来源：https://stackoverflow.com/questions/14295426/c-sizeof-char-pointer