what is the output of this code
Hi all,
the output is 6 and 4Code:#include<stdio.h> int main() { char p[]="hello"; printf("sizeof(p)=%d\n",sizeof(p)); printf("sizeof(p+1)=%d\n",sizeof(p+1)); return 0; }
Can someone explain why is the output of sizeof(p+1)=4
Regards and Thanks
C_Enthusiast
p by itself is an array.
p+1 is an expression. Arrays in expressions become "a pointer to the first element". So what you end up with is sizeof(char*)
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Take a look here,
sizeof - Wikipedia, the free encyclopedia
