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I saw a weird type of program here.
int main() { int s[]={3,6,9,12,18}; int* p=+s; } Above program tested on GCC and Clang compilers and working fine on both compilers.
I curious to know, What does int* p=+s; do?
Is array s decayed to pointer type?
885
so p= (int *) &i means p is storing the address of i which is of type char but you have type cast it, so it's fine with that. now p is point to i. *p = 123455; // here you stored the value at &i with 123455. when you'll print these value like. print (*p) // 123455.
A pointer to a pointer is a form of multiple indirection, or a chain of pointers. Normally, a pointer contains the address of a variable. When we define a pointer to a pointer, the first pointer contains the address of the second pointer, which points to the location that contains the actual value as shown below.
The strlen() Function The strlen() accepts an argument of type pointer to char or (char*) , so you can either pass a string literal or an array of characters. It returns the number of characters in the string excluding the null character '\0' .
Built-in operator+ could take pointer type as its operand, so passing the array s to it causes array-to-pointer conversion and then the pointer int* is returned. That means you might use +s individually to get the pointer. (For this case it's superfluous; without operator+ it'll also decay to pointer and then assigned to p.)
(emphasis mine)
The built-in unary plus operator returns the value of its operand. The only situation where it is not a no-op is when the operand has integral type or unscoped enumeration type, which is changed by integral promotion, e.g, it converts char to int or if the operand is subject to lvalue-to-rvalue, array-to-pointer, or function-to-pointer conversion.
139
Test this:
#include <stdio.h> int main(){ char s[] = { 'h', 'e', 'l', 'l', 'o' , ' ', 'w', 'o', 'r', 'l', 'd', '!'} ; printf("sizeof(s) : %zu, sizeof(+s) : %zu\n", sizeof(s), sizeof(+s) ) ; } On my PC (Ubuntu x86-64) it prints:
sizeof(s): 12, sizeof(+s) : 8 where
12 = number of elements s times size of char, or size of whole array 8 = size of pointer If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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