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REALLY EASY I don't know why this won't enter the for loop

I keep getting compiling errors on this and I'm about ready to throw my first right through the screen of this laptop.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
 
int main( void )
{
	int quantity = 1;
	int i = 1;
	srand(time(NULL));
 
	printf( "How many Quick Picks do you want?\n" );
	scanf( "%d\n", &quantity );
 
	for ( i = 1; i <= quantity i++ ) 
	{
		printf( "Quick Pick: %d\n", 1 + (rand() % 54));
	}
			
	return 0;  /* program executed successfully */
}

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Doomtomb
Asked:
Doomtomb
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1 Solution
 
sunnycoderCommented:
for ( i = 1; i <= quantity; i++ )
note ; after quantity
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DoomtombAuthor Commented:
Jeeze, I knew it was something dumb like that. I have attached what it is right now but for some reason, it won't enter the for loop. It prints asking the user, then you type a value like 2 and enter and nothing happens.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
 
int main( void )
{
	int quantity;
	int i;
	srand( time( NULL ) );
 
	printf( "How many Quick Picks do you want?\n" );
	scanf( "%d\n", &quantity );
 
	for ( i = 1; i <= quantity; i++ ) {
		printf( "Quick Pick: %d\n", 1 + (rand() % 54));
	}
			
	return 0;  /* program executed successfully */
}

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sunnycoderCommented:
>scanf( "%d\n", &quantity );

scanf has its quirks ... remove the \n in the format string ...
preferred method of inputting is to use fgets to read in a line and then use atoi/strtol to convert.
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DoomtombAuthor Commented:
Works great. My programming teacher forces us to use scanf. It sucks. I HATE THIS!!!!! I GOT SO MAD!!!!!!! THANK YOU!!!!!! *ANGRY VOICE*
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