Programming in C (programme to visualize the 10 next number of n)

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Hi,
I'm making a programme to visualize the next 10 number of n. I know, is very easy but i'm new in this.

Here is my programme. I can't visualize the 10 numbers after n. I always get one number (n+10). For exemple, if a put n=5, i need to visualize 6,7,8,9,10,....,15.
Why i always get just 15?

C++

#include<stdio.h>
void numeros_sumados(int n)
{
int i=0;
int M;

	for (i=n+1; n+10;i++)
	{

		fprintf(stdout,"%d\n",i);
		
	}
		fprintf(stdout,"\n");
}

int main (void)
{
int n;

	fprintf(stdout,"Tapez le numero n:\n");
	fscanf(stdin,"%d",&n);
	numeros_sumados(n);
}

Thanks

Posted 15-Nov-13 8:53am

alonetmr

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Ron Beyer · Answer 1 · 2013-11-15T09:07:00

Solution 1

Take a look at this line:

C++

for (i=n+1; n+10;i++)

Remember how a for-loop works?

The format is:

for (initial condition; exit condition; step amount)

So you have the initial condition set (i = n + 1)
But the exit condition is just n+10, which evaluates to true (greater than 0), and only loops once.

You need something like this:

C++

for (i=n+1; i < n + 10; i++)

Which will loop until i is greater than or equal to n+10.

Posted 15-Nov-13 9:07am

Ron Beyer

Comments

alonetmr 15-Nov-13 15:25pm

Hi, thanks for you answer but it doesn't work. I always get one value and the loop never end.

Albert Holguin 15-Nov-13 16:23pm

This should work fine unless you did something else wrong.

Albert Holguin 15-Nov-13 16:22pm

+5... I like that you took the time to explain his problem (vs just doing it right for him)

Andreas Gieriet 15-Nov-13 18:01pm

The code is right, the explanation is wrong, sorry.
The condition is "loop while the condition is true" - the condition is the *opposite* of an exit condition.
for(init;cond;inc) { ... } is equivalent to init; while(cond) { ... inc; }
Cheers
Andi

Richard MacCutchan 16-Nov-13 4:48am

Actually, the test condition of n+10 means it will loop for ever.

OriginalGriff · Answer 2 · 2013-11-15T09:11:00

Solution 2

Simple: your termination condition in the for loop isn;t a condition - it's a non-zero value, which is true as far as C is concerned...
Try changing:

C++

for (i=n+1; n+10;i++)

To:

C++

for (i=n+1; i < n+10;i++)

And see what happens!

Posted 15-Nov-13 9:11am

OriginalGriff

Comments

alonetmr 15-Nov-13 15:25pm

Hi, thanks for you answer but it doesn't work. I always get one value and the loop never end.

OriginalGriff 15-Nov-13 15:58pm

Copy and paste *exactly* the code you are trying - complete with the change I recommended.

Because I think you typed it wrong... :laugh:

alonetmr 15-Nov-13 16:03pm

I followed your suggestion. I put n=2 and I got this:
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12^C

Why? is always the same number. thank you

By the way, here is the programme with the modification:

#include<stdio.h>
void numeros_sumados(int n)
{
int i=0;

for (i=n+1; i < n+10;i++)
{

fprintf(stdout,"%d\n",i);

}
//fprintf(stdout,"\n");
}

int main (void)
{
int n;

fprintf(stdout,"Tapez le numero n:\n");
fscanf(stdin,"%d",&n);
numeros_sumados(n);
}

OriginalGriff 15-Nov-13 16:18pm

No it isn't - that won't even compile!
There is no 'n' in scope of your method: your parameter is called 'p'
So now look at the version that compiles... :laugh:

Albert Holguin 15-Nov-13 16:27pm

Guessing you and Ron responded at the same time... +5 anyway, :)

alonetmr 15-Nov-13 16:36pm

Ok, now it works. thank you )

OriginalGriff 15-Nov-13 16:39pm

You're welcome!