Literal string initializer for a character array
arrays, c, c99, pointers, string
Solution
The three exceptions where an array does not decay into a pointer are the following:
Exception 1. — When the array is the operand of `sizeof`.
int main()
{
int a[10];
printf("%zu", sizeof(a)); /* prints 10 * sizeof(int) */
int* p = a;
printf("%zu", sizeof(p)); /* prints sizeof(int*) */
}
Exception 2. — When the array is the operand of the `&` operator.
int main()
{
int a[10];
printf("%p", (void*)(&a)); /* prints the array's address */
int* p = a;
printf("%p", (void*)(&p)); /*prints the pointer's address */
}
Exception 3. — When the array is initialized with a literal string.
int main()
{
char a[] = "Hello world"; /* the literal string is copied into a local array which is destroyed after that array goes out of scope */
char* p = "Hello world"; /* the literal string is copied in the read-only section of memory (any attempt to modify it is an undefined behavior) */
}
Problem
In the following rules for the case when array decays to pointer: An lvalue [see question 2.5] of type array-of-T which appears in an expression decays (with three exceptions) into a pointer to its first element; the type of the resultant pointer is pointer-to-T. (The exceptions are when the array is the operand of a sizeof or & operator, or is a literal string initializer for a character array.) How to understand the case when the array is "literal string initializer for a character array"? Some example please. Thanks!