modify structure array pointer in the function

[s88]

Howdy:
the follows is my program, I wanna change my structure array
pointer in the function "testfunc", but I fail..., I also try to call
the testfunc by reference, but the compiler says "test6.c:35: error:
incompatible type for argument 1 of `testfunc'". Can I make my purpose
in C?
and how?


typedef struct xxxx *xxxx_ptr;

typedef struct xxxx{
int just;
long for_the;
char *test;
}xxxx;

void testfunc(xxxx_ptr ptr){
ptr++;
}

int main(void){
xxxx XXXX[10];
xxxx_ptr ptr = &XXXX[0];
printf("%x\n",ptr);
testfunc(ptr);
printf("%x\n",ptr);
return 0;
}

Thank you all.
Dave.

 

[pete]

Howdy:
the follows is my program, I wanna change my structure array
pointer in the function "testfunc", but I fail..., I also try to call
the testfunc by reference, but the compiler says "test6.c:35: error:
incompatible type for argument 1 of `testfunc'". Can I make my purpose
in C?
and how?

typedef struct xxxx *xxxx_ptr;

typedef struct xxxx{
int just;
long for_the;
char *test;
}xxxx;

void testfunc(xxxx_ptr ptr){
ptr++;
}

int main(void){
xxxx XXXX[10];
xxxx_ptr ptr = &XXXX[0];
printf("%x\n",ptr);
testfunc(ptr);
printf("%x\n",ptr);
return 0;
}

Every oportunity to use typedef,
isn't necessarily a good one.

/* BEGIN new.c */

#include <stdio.h>

struct my_struct {
int just;
long for_the;
char *test;
};

void testfunc(struct my_struct **ptr)
{
++*ptr;
}

int main(void)
{
struct my_struct array[10];
struct my_struct *ptr = array;

printf("%p\n", (void *)ptr);
testfunc(&ptr);
printf("%p\n", (void *)ptr);
return 0;
}

/* END new.c */

 

[rohin.koul]

On the contrary when I copied your code into old Turbo C its compiled
and run just fine.
So I guess its a compiler issue

 

[Barry Schwarz]

Howdy:
the follows is my program, I wanna change my structure array
pointer in the function "testfunc", but I fail..., I also try to call
the testfunc by reference, but the compiler says "test6.c:35: error:
incompatible type for argument 1 of `testfunc'". Can I make my purpose
in C?
and how?


typedef struct xxxx *xxxx_ptr;

typedef struct xxxx{
int just;
long for_the;
char *test;
}xxxx;

void testfunc(xxxx_ptr ptr){

C passes arguments by value. The ptr is this function is a copy of
the argument value in the calling statement.

ptr++;

This updates the copy.

}

The copy is destroyed as part of the process of exiting the function.
The updated value no longer exists.

int main(void){
xxxx XXXX[10];
xxxx_ptr ptr = &XXXX[0];
printf("%x\n",ptr);

%x expects an unsigned int. You are passing a pointer. This invokes
undefined behavior.

The only portable way to print a pointer value is to use %p and cast
the value to a void *.

testfunc(ptr);

The value of an argument cannot be changed by the called program.

printf("%x\n",ptr);
return 0;
}

The two usual solutions are:

Have the function return the updated value and call the function
in an assignment statement that assigns the new value to a suitable
variable.

Pass the address of the variable to be updated and let the
function update the variable by dereferencing the address.


<<Remove the del for email>>

 

[Keith Thompson]

the follows is my program, I wanna change my structure array
pointer in the function "testfunc", but I fail..., I also try to call
the testfunc by reference, but the compiler says "test6.c:35: error:
incompatible type for argument 1 of `testfunc'". Can I make my purpose
in C?
and how?


typedef struct xxxx *xxxx_ptr;

typedef struct xxxx{
int just;
long for_the;
char *test;
}xxxx;

void testfunc(xxxx_ptr ptr){
ptr++;
}

int main(void){
xxxx XXXX[10];
xxxx_ptr ptr = &XXXX[0];
printf("%x\n",ptr);
testfunc(ptr);
printf("%x\n",ptr);
return 0;
}

Are you sure that's the actual code you fed to the compiler? When I
compiled it with gcc, I didn't get any diagnostics. When I
upped the warning level (gcc -ansi -pedantic -Wall -W -c tmp.c), I got:

tmp.c: In function `main':
tmp.c:16: warning: implicit declaration of function `printf'
tmp.c:16: warning: unsigned int format, xxxx_ptr arg (arg 2)
tmp.c:18: warning: unsigned int format, xxxx_ptr arg (arg 2)

You need a "#include <stdio.h>" to make the printf() calls legal.
Also, the correct format for a pointer is "%p", not "%x":

pirntf("%p\n", (void*)ptr);

(Since you got an error message on line 35, what you posted apparently
is shorter than the actual code.)

 

[Blair Craft]

Try this:
/* start */
typedef struct xxxx *xxxx_ptr;

typedef struct xxxx{
int just;
long for_the;
char *test;
}xxxx;

void testfunc(xxxx_ptr **ptr){
(*ptr)++;
}

int main(void){
xxxx XXXX[10];
xxxx_ptr ptr = XXXX;
printf("%x\n",ptr);
testfunc(&ptr);
printf("%x\n",ptr);
return 0;
}
/* end */