Thread: C newbie needing help.

Code:

for ( x = 0; x < sizeof(numbers) ; x++ )

Are you sure that the size allocated at compile time and sizeof(numbers) same?
Check for yourself by adding a printf statement

Code:

.
.
printf("Size of my array : %d", sizeof(numbers));
	for ( x = 0; x < sizeof(numbers) ; x++ ) {
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Originally Posted by Enzo

..but it would be nice if I didn't have to put that 0.

Tell us how exactly would you want to terminate the input.

Originally Posted by zalezog

Are you sure that the size allocated at compile time and sizeof(numbers) same?
Check for yourself by adding a printf statement

You made me notice something quite surprising to me there. I didn't know it could be different "at compile time" and I'm really not sure why does it happen. I used the printf and it said sizeof(numbers) = 200, when it should have been 50.

About the number input, I wanted to be able to finish when pressing enter. I managed to do this by using the functions "strtol", "fgets" and doing some other stuff a bit more complicated. However, I'm still confused about what MK57 suggested me to do in regards with the scanf function. Because when I wait until scanf < or != 1, it doesn't happen. It just keeps inputing characters. When does scanf not read a character sucessfully? How can I make the input stop, when hitting enter with the scanf funtion, if it just keep inputing? Also why is sizeof(numbers) different at compile time?

Thank you for you time and for answering my questions! Your help is greatly appreciated!

Code:

  for ( x = 0; scanf("%d",&numbers[x]) == 1 ; x++ )
        ;
    printf ( "\n" );

In case my code is wrong, I found this one in wikipedia.org, which gives me the same results

Code:

int n;
    while (scanf("%d", &n) == 1){
        printf ("%d\n", n);

thank you again!

The problem with scanf() is that's it's basically mindless. If you put in %d for the format specifier, it will happily wait for that number - you can press enter until the cows come home, basically.

There are probably ways around that problem with scanf() specifiers, but scanf() just doesn't strike me as being good for flexible user input.

I would do it something like this:

Code:

#include <stdio.h>
#include <stdlib.h>

int main () {
    char buff[20];
    int n, x = 0;
    int numbers[10];

    do {
      fgets(buff, sizeof(buff), stdin);
      numbers[x++] = atoi(buff);
    }while(buff[0] != '\n'); 
      
    --x;  
    printf ( "\n You entered %d numbers", x );

    for ( n = 0; n < x; n++ ) {
        printf ( "\n%d", numbers[n] );
    }

    printf("\n\n\t\t\t     press enter when ready");
    x = getchar();

    return 0;
}

Try that and see what you think.

Hello !
Try this technique. It works and fulfills your requirement

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main () {

int x = 0 ;
int cnt;
int numbers[50];
char str[100];
char *ptr ;

// fgets(str,100,stdin);
// or
gets(str);
ptr = strtok (str," ");
while( ptr != NULL )
{
numbers[x++] = atoi(ptr);
ptr = strtok(NULL," ") ;
}
printf ( "\n" );

for ( cnt = 0; cnt < x; cnt++ ) {
printf ( "\t%d", numbers[cnt] );
}
return 0;
}