how can i do it?

linuxanswer · 11-03-2003, 06:14 PM

#include <stdio.h>
#define SizeA 30

main() {
int i,k,j;
int argv1[SizeA], argv2[SizeA] ;

for(j=0;j<SizeA;j++) {

printf("Array uno inizializzato con i primi 30 numeri\n\n");
break;
}
for( i=1, k=30; i<(SizeA/2), k>(SizeA/2); i++, k--) {

argv2[SizeA]=argv2[i,k];

printf("%d,%d;", i, k);

}

return(1);

}

I want put i and k like elements of argv2 so at last i'll print 1,30;2,29;3,28;4,27;5,26;6,25;7,24;8,23;9,22;10,21;11,20;12,19;13,18;14,17;15,16;

how can i do it?

linuxanswer · 11-03-2003, 06:15 PM

that 's to say, it's correct argv2[SizeA]=argv2[i,k];?

djtoltz · 11-03-2003, 06:21 PM

This is a "C" programming question, not really a Linux question. You've got some issues with your C syntax, for starters. However, you don't need the arrays for what you are trying to do. All you really need is two variables that start at min and max and meet in the middle.

Something like the following should work...

#include <stdio.h>

int main() {
int i, j;
for (i=1, j=30; i < j; i++, j--)
{
printf("%d,%d,",i,j);
}
return 0;
}

djtoltz · 11-03-2003, 06:25 PM

If you are trying to fill in an array with the numbers of your sequence, the syntax would be something like this;

/* initialize a_index to zero before this */
argv2[a_index++]=i;
argv2[a_index++]=j;

... this would put i and j into the array at the offset <a_index> which would continue to be bumped up as elements were added.

LogicG8 · 11-03-2003, 07:28 PM

I put some comments around your code to help you out.

Code:

#include <stdio.h>
#define SizeA 30	/* It is convention to put constants all in CAPS */

main() {
	int i,k,j;
	/* 
	* argv1 is unused and it is a convention to use
	* argv to refer the the values of
	* the arguments passed to your program
	* Argument Values hence argv
	*/
	int argv1[SizeA], argv2[SizeA] ; 

	/*
	* This loop only prints your string
	* once this is is silly
	*/
	for(j=0; j<SizeA; j++) {
		printf("Array uno inizializzato con i primi 30 numeri\n\n");
		break;
	}
	for( i=1, k=30; i<(SizeA/2), k>(SizeA/2); i++, k--) {
		/* 
		* SizeA is one index outside of the array and 
		* it makes no sense to loop assigning the same
		* variable to different values
		*/
		argv2[SizeA]=argv2[i,k]; 
		printf("%d,%d;", i, k);
	}
	return (1); /* Zero is customary for indicating success */
}

My version would look like

Code:

#include <stdio.h>
#define SIZE_A 30

int main(int argc, char *argv[])
{
        int i;
        int array0[SIZE_A];
        int array1[SIZE_A];

        /*
        * I don't know why I do this. Your program
        * just says that is what it is doing so I
        * actually do it.
        */
        printf("Initialize array with first 30 numbers\n");
        for (i = 0; i < SIZE_A; i++) {
                array0[i] = i;
        }


         for (i = 0; i < SIZE_A; i += 2) {
                array1[i] = (i+2) / 2;
                array1[i+1] = SIZE_A - (i/2);
                printf("%d,%d;", array1[i], array1[i+1]);
        }

        printf("\n");
        return 0;
}

Here you go and good luck.