problem with int64 in rh fdeora 4 ...

mohtashami · 11-05-2005, 05:30 AM

Hi ALL.
I have a problem in call reference with int64 (in redhat fedora core 4 with gcc 4.0).
see this simple code:

//test1.c -->
#include <stdio.h>
#define ull unsigned long long int

void foo(ull a){
printf("a= %llx\n",a);
}
int main(void){
ull a=0x123456789abcde;
foo(a);
return 0;
}
//test1.c <--

this code is compiled with this warning :
test.c: In function ‘main’:
test.c:8: warning: integer constant is too large for ‘long’ type

but output file executes with correct output.

but see this code
//test2.c -->
#include <stdio.h>
#define ull unsigned long long int
void foo(ull& a){ //changed to call by reference
printf("a= %llx\n",a);
}
int main(void){
ull a=0x123456789abcde;
foo(a);
return 0;
}
//test2.c <--

this code generates this errors :

test.c:4: error: syntax error before ‘&’ token
test.c: In function ‘foo’:
test.c:5: error: ‘a’ undeclared (first use in this function)
test.c:5: error: (Each undeclared identifier is reported only once
test.c:5: error: for each function it appears in.)
test.c: In function ‘main’:
test.c:8: warning: integer constant is too large for ‘long’ type

I want to use an unsigned int 64 with call reference.this code has no error with redhat 7.3 (gcc 2.96).
please help me ....

spooon · 11-05-2005, 06:00 AM

It works if you put "ull" or "ULL" (for unsigned long long) at the end of the constant.

mohtashami · 11-05-2005, 06:41 AM

qoute -->
It works if you put "ull" or "ULL" (for unsigned long long) at the end of the constant.
<-- unqoute

the warning is not my problem !
my problem is error in second code.

mohtashami · 11-05-2005, 07:05 AM

By adding 'll' to end of definition of a, first code compiles with no warning but second code has error.

spooon · 11-05-2005, 05:26 PM

Oh yeah, I'm sorry.

There are no references in C. They are only available in C++. It's equivalent to using pointers like this:

Code:

#include <stdio.h>
#define ull unsigned long long int
void foo(ull* a){ //changed to call by reference
printf("a= %llx\n",*a);
}
int main(void){
ull a=0x123456789abcdeull;
foo(&a);
return 0;
}