how to assign a fix value to int in c

Question

I am trying to achieve the following in C. I have ten objects, each of them have a different number assign to them and they have their own value as well .For example 1 = 1000. While the user input 1, it should reply the value of 1 which is 1000. where do i need to defied the object? in INT or using char? Also it seems that i can't int = an number why is that?

Here is the unfinished code that I am working with.

#include <stdio.h>
void main()
{
    int "1" ;

    "1= 1000";
    scanf( "%d ",&number2 );
    number2 = c;
    if ( c == 1)
        printf(,1)
}

Answer 1

There are some problem in this code :

1) According to the standard you shouldn't use

void main()

But either

int main()
int main(void)
int main(int argc, char *argv[])

Also never forget

return 0;

Sample program :

#include <stdio.h>

int main(void) {       
    /* Do stuff */        
    return 0;    
}

If you wish to read more information about that, click here.

2)

int "1" ;

That's not a correct way to declare a variable. You should have write :

int variable_name = 1;

Moreover, if you want to declare an array of 10 integer elements you have to write :

int array[10] = {....}

3)

"1= 1000";

I guess you want to overwrite the value of the variable prior declared with "1". Following my example given before :

variable_name = 1000;

4)

scanf( "%d ",&number2 );

number2 = c;

You didn't declare either the variable number2 nor c.

5)

printf(,1)

That's not how you use printf. My guess here is that you tried to print :

int "1" ;

Following my example you can achieve this doing:

printf("%d", variable_name);

EDIT : My advices given above are corrects. Perhaps you was looking for something like that :

#include <stdio.h>

int main(void){

        int array[10] = {1000,2000,3000,4000,5000,6000,7000,8000,9000,10000}, user_choiche = 0, sum = 0, i = 0;

        do
        {
                printf("\nEnter a valid position (0 <= n <= 10) : ");
                scanf("%d", &user_choiche);
        }
        while(user_choiche < 0 || user_choiche > 10);

        printf("\nThe value stored in the %d position of the array is : %d\n\n", user_choiche, array[user_choiche]);

        for(i = 0; i < 10; i++)
             sum += array[i];

        printf("\nSum is %d.\n\n", sum);


return 0;

}

Input :

Enter a valid position (0 <= n <= 10) : -2

Enter a valid position (0 <= n <= 10) : 2

Output :

The value stored in the 2 position of the array is : 3000

Sum is 55000

Answer 2

You should honestly put more effort before posting something like that. You and I have got something in common, we are both noobs, yet that is not an excuse for not putting the right ammount of effort beforehand. Do some research before asking a question, most of the times, you will find out that it is in your power to solve it, the secret is to persevere and not take things for granted. Here, this will help for starters:

http://www.catb.org/~esr/faqs/smart-questions.html (we are not a help desk for your project!)

Answer 3

Try this :

int arraynumbers[11] = {0, 1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10000};
int input;
scanf("%d",&input);
printf("%d", arraynumbers[input]);