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adding from 0 to a number that will be entered by the user .

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mrayoub

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adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

hi,

well this is my first post on this great forum, hope to be a good boy

i m writing a programm that adds numbers from 0 to any numer entered by the user .

for Example :

- if the user enter 4, the sum will be 10 = 0+1+2+3+4.
- if 10 --> the sum is : 1+2+3+4...+10 = 64 .

i make the code like that :

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#include <stdio.h>
#include <stdlib.h>
 
int main()
{
    int number;
    int sum;
    int i;
 
    printf("please enter the end number : \n" );
    scanf("%i", &number);
 
    for ( i = 0; i <= number; i++ ) {
 
        sum += i;
 
    }
    printf("the sum is : %i\n", sum);
    return 0;
 
 
}
#include <stdio.h>
#include <stdlib.h>

int main()
{
    int number;
    int sum;
    int i;

    printf("please enter the end number : \n" );
    scanf("%i", &number);

    for ( i = 0; i <= number; i++ ) {

        sum += i;

    }
    printf("the sum is : %i\n", sum);
    return 0;


}

the result :

the sum is : 2293586 when i enter 4

i don t know why .

any ideas how to make this work

Thanks.

Last edited by mrayoub; Jun 27th, 2009 at 12:53 pm.

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mathematician mathematician is offline

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Re: adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

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

int main()
{
    int number;
    int sum;
    int i;
 
    printf("please enter the end number : \n" );
    scanf("%u", &number);
 
    for ( i = 0, sum = 0; i <= number; i++ ) {
        sum += i;
    }

    printf("the sum is : %i\n", sum);

    return 0;
}

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mrayoub

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Re: adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

thank you very much mathematician for this quick answer but i have small question why did you use %u in the place of %i i mean it works with %i too .

thanks again

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Re: adding from 0 to a number that will be entered by the user .

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Jun 27th, 2009

its just

sum=sum+i

hehehe

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Re: adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

for ( i = 0; i <= number; i++ ) {
sum =sum+ i;
}

try this code...

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Re: adding from 0 to a number that will be entered by the user .

-1

Jun 27th, 2009

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Originally Posted by mrayoub

thank you very much mathematician for this quick answer but i have small question why did you use %u in the place of %i i mean it works with %i too .

thanks again

Well firstly, unless it is compiler specific, I have never seen %i before. A decimal integer is usually %d.

Secondly, given what you are doing, you presumably want the user to enter a positive (unsigned) integer (%u).

Last edited by mathematician; Jun 27th, 2009 at 2:10 pm.

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mrayoub

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Re: adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

thanks james14 it works too ( well i m a beginner i coudn t figure it out

)

thanks

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Originally Posted by mathematician

Well firstly, unless it is compiler specific, I have never seen %i before. A decimal integer is usually %d.

Secondly, given what you are doing, you presumably want the user to enter a positive (unsigned) integer (%u).

well we always do this, we work just with %i if its an integer and it works

but the idea to use %ufor the positive numbers is great

thank you mathematecian for solving this .

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mrayoub

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Re: adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

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Originally Posted by james14

for ( i = 0; i <= number; i++ ) {
sum =sum+ i;
}

try this code...

well if sum doesn t initialised it wouldnt work !

for(i = 0, sum = 0; i <= number; i++)

will be correct

thanks

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Re: adding from 0 to a number that will be entered by the user .

Jun 27th, 2009

mrayoub:

%i and %d are equivalent. both of them specify the format for a signed decimal integer, and either are acceptable to use.

%u, as you have seen, is for unsigned decimal integer

however, if you have declared a variable of type int then you should not print it as %u, an unsigned int. That is a bad practice to get into... because while it may in many cases appear to work correctly, your integer may very well have a valid negative value , and the %u format will print an incorrect value.

if you want the variable to be unsigned, then make it unsigned by declaring it so: unsigned number;

.

Last edited by jephthah; Jun 27th, 2009 at 8:57 pm.