warning: taking address of temporary - c++

Question

i have the following code snippet
if((compare( & key,& ALL))
All is #defined as
#define ALL (some_structure){"ALL", 3}
in g++ version
g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-46)
compiled without any warnings
but in g++ version
g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-4)
it is showing warning: taking address of temporary
even though it is a warning.the whole compile message shows aroung 3 million plus messages.It is a legacy code.is there any way to suppress this warning?
Edit :
Below is the sample code
#include <iostream>
using namespace std;
typedef struct test { char * name; int id; }tsTest;
void print(tsTest * T) {
cout<<endl<<"Name :"<<T->name<<endl<<"ID :"<<T->id<<endl; }
int main()
{
print(&(tsTest){"nagaraj",7});
return 0;
}
compiled fine in g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-46)
But in g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-4)
temadd.cpp: In function ‘int main()’:
temadd.cpp:18: warning: deprecated conversion from string constant to ‘char*’
temadd.cpp:18: warning: taking address of temporary

Answer 1

As a workaround, you can implement operator& of the structure. Then you would not take address of the temporary, that is undefined behaviour (UB), but calling function of temporary object, returning pointer to itself.
As temporary will live as long as function call, you will not encounter any UB unless you store the pointer and dereference it after leaving the function
For your example it would be like:
typedef struct test {
const char * name;
int id;
test* operator&() {
return this;
}
}tsTest;

Answer 2

It's not a warning you should suppress. The problem with taking the address of a temporary is that when that temporary goes out of scope your application then exhibits undefined behaviour.
Do you have any reason for not defining ALL as a const global? You could safely take its address at any time without any UB:
struct allstruct
{
char *allname;
int somenumber;
};
const allstruct ALL { "ALL", 3 };