Assume there's function that get int * parameter.
void foo(int *x)
{
}
If I want to call this function without creating an int variable
int main()
{
foo(&1);
return 0;
}
compilation fails and I get this error message:
lvalue required as unary ‘&’ operand
Why 1 has no address? Isn't that stored in stack ?
Why 1 has no address? Isn't that stored in stack ?
No - it is just a value.
You need an object to take its address (reference). A constant expression is not such an object.
You can use compound literals to create temporary object:
void foo(int *x)
{
printf("%d\n", *x);
}
int main(void)
{
foo( (int[]){1} );
foo( &(int){1} );
}
https://godbolt.org/z/xvdbTzn3Y
but string constants have addresses what is the difference ?
They are not constant expressions only string literals. String literal is an array of char``[C] or const char``[C++]. String literals cannot be modified (it invokes undefined behabior) . It is an object and you can take the address of it:
void foo(char (*x)[])
{
printf("'%s'\n", *x);
}
int main(void)
{
foo(&"dsfgdfsgdfg");
}