C: structs and pointers question

kpachopoulos · 03-25-2006, 08:11 AM

Hi,
how can i get the print function to print the the argv[i] instead from their addresses? I tried changes in the print function, like "r.p1.(*x)", "r.p1.*x" and "*r.p1.x", which didn't work...

Code:

#include <stdio.h>

struct point {
	int* x;
	int* y;
} a_point, b_point;

struct rect {
	struct point p1;
	struct point p2;	
} a_rect;

void print(struct rect r)
{
	printf("%i || %i \n", r.p1.x, r.p1.y);
	printf("%i || %i \n", r.p2.x, r.p2.y);	
}

main(int arc, char *argv[])
{
	
	a_point;
	a_point.x=argv[1];
	a_point.y=argv[2];
	
	b_point;
	b_point.x=argv[3];
	b_point.y=argv[4];
		
	a_rect.p1=a_point;
	a_rect.p2=b_point;
	print(a_rect);
	
}

graemef · 03-25-2006, 08:58 AM

argv is a character array, so you will need to convert it from the string to an integer before assigning it. look at the atoi function for some assistance

jtshaw · 03-25-2006, 08:33 PM

As an aside.. (assuming your on linux) I'd recommend strtol or strtoll over atoi....

long strtol (char *str, char *endptr, int base);
long long strtoll (char *str, char *endptr, int base);

base can be any number from 2 to 36.. endptr is a pointer to the place in the string you want it to stop, just pass in null if you want it to stop either at the first invalid char or the end of the string.

Example:
a_point.x = strtol(argv[1],NULL,10);

ta0kira · 03-27-2006, 05:54 PM

The * goes at the front: '*r.p1.x'. 'r.p1.*x' means that some variable 'x' in the current scope is a pointer to either a member function or member data, whereas 'r.p1.x' is the fully-qualified pointer that you need to dereference.

In your particular case, however, you are trying to convert members of the 'argv' array to integers, however they are taken off the command line as strings. You are looking for something like 12345, whereas the data you are looking at is "12345" (an array of 5 characters.) You need to convert them to integers, which can be done with atoi.
ta0kira