C sizeof char pointer

Question

Why is size of this char variable equal 1?

int main(){

char s1[] = \"hello\";

fprintf(stderr, \"(*s1) : %i\\n\", sizeof(*s1) )    // prints out 1

}


        

Answer 1

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.

Answer 2

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.

Answer 3

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])

Answer 4

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.

Answer 5

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]);