Help w/ C Programming Homework - Create Array & Print All Inputs

[doktorwho]

Homework Statement

I am fairly new to C programming. I need to write a code that creates an array of max 300 numbers, ask the user to input the numbers and then print that array.

Homework Equations

3. The Attempt at a Solution [/B]
Here is my code

#include <stdio.h>

void main() {
    int numbers[300];
    int i, n;

    printf("Enter the size of your array: \n");
        scanf("%d", &n);

    printf("Enter the numbers: \n");
    for (i = 0; i < n; ++i);
        scanf("%d", &numbers[i]);

    printf("Now your array will be printed: \n");
    for (i = 0; i < n; ++i);
        printf("%d ", numbers[i]);
      
        return 0;
}

The problem is that it only prints the first number inputed. How do i fix this?

 

[Drakkith]

Check your syntax for your for loops. As soon as I corrected those, the code appeared to work fine.

 

[Mark44]

for (i = 0; i < n; ++i);
    scanf("%d", &numbers[i]);

As @Drakkith said, the syntax of your for loops is wrong.
Your indentation is misleading. All the for loop does is count from 0 through n - 1, and then it's done. Next, it attempts to take input from the user and store it at index n in your array.

Yours is a common mistake among new C (and related languages) programmers.

 

[doktorwho]

Check your syntax for your for loops. As soon as I corrected those, the code appeared to work fine.

for (i = 0; i < n; ++i);
    scanf("%d", &numbers[i]);

As @Drakkith said, the syntax of your for loops is wrong.
Your indentation is misleading. All the for loop does is count from 0 through n - 1, and then it's done. Next, it attempts to take input from the user and store it at index n in your array.

Yours is a common mistake among new C (and related languages) programmers.

#include <stdio.h>
void main() {
    int numbers[300];
    int i, n;

    printf("Enter the size of your array: \n");
    scanf("%d", &n);

    printf("Enter the numbers: \n");
    for (i = 0; i < n; i++)
    {
      scanf("%d", &numbers[i]);
    }

    printf("Now your array will be printed: \n");
      for (i = 0; i < n; i++)
    {
      printf("%d ", numbers[i]);
    }
       
    return 0;
}

So when i put ; at the end i basically just said that my for loop only counts and does nothing else. Could you explain how is it now corrected? I get the right outputs now.

 

[Mark44]

So when i put ; at the end i basically just said that my for loop only counts and does nothing else.

Yes, exactly.

Could you explain how is it now corrected? I get the right outputs now.

The syntax of a for loop is

for( <init expression>; <test expression>; <update expression>) <statement>;

<statement> can be empty, which is what you had before.
Basically, you had an empty for loop that just counted, and following it a block statement--the call to scanf() surrounded by braces. As far as the compiler was concerned, there was no connection between the for loop and the following block.

 

[Drakkith]

So when i put ; at the end i basically just said that my for loop only counts and does nothing else. Could you explain how is it now corrected? I get the right outputs now.

In C, semicolons terminate statements, so putting a semicolon right after the first for loop line prevents it from executing anything underneath it. That's why you leave the semicolon out and add the curly braces. That way the compiler knows that the for loop is to check the conditional statements first, then execute anything inside the brackets before looping back through.

 

[Mark44]

In C, semicolons terminate statements, so putting a semicolon right after the first for loop line prevents it from executing anything underneath it. That's why you leave the semicolon out and add the curly braces.

The braces aren't required in all cases.

for (i = 0; i < 10; i++)
    printf ("Hello\n");

However, if the body of the for loop consists of two or more statements, the braces are required.

for (i = 0; i < 10; i++)
    printf ("Hello\n");
    printf("World");

This loop, which has misleading indentation, would print Hello 10 times, and would then print World once, as it is not part of the loop body. Humans might be fooled by the indentation, but the compiler isn't.

So, even though the braces aren't required if the loop body consists of only one statement, it's a good idea to include them anyway, in the possibility that you come back and add another statement to what you think is the loop body, but actually isn't.

 

[doktorwho]

The braces aren't required in all cases.

for (i = 0; i < 10; i++)
    printf ("Hello\n");

However, if the body of the for loop consists of two or more statements, the braces are required.

for (i = 0; i < 10; i++)
    printf ("Hello\n");
    printf("World");

This loop, which has misleading indentation, would print Hello 10 times, and would then print World once, as it is not part of the loop body. Humans might be fooled by the indentation, but the compiler isn't.

So, even though the braces aren't required if the loop body consists of only one statement, it's a good idea to include them anyway, in the possibility that you come back and add another statement to what you think is the loop body, but actually isn't.

And to print the Hello World 10 times with the same code i do:

for (i = 0; i < 10; i++)
    printf ("Hello\n")
    printf("World");

which makes the one loop end after the world?

 

[Drakkith]

And to print the Hello World 10 times with the same code i do:

for (i = 0; i < 10; i++)
    printf ("Hello\n")
    printf("World");

which makes the one loop end after the world?

Since your loop consists of more than 1 statement, you need curly braces.

 

[Mark44]

And to print the Hello World 10 times with the same code i do:

for (i = 0; i < 10; i++)
    printf ("Hello\n")
    printf("World");

which makes the one loop end after the world?

No. The second printf() statement is not part of the for loop.
The code above is exactly the same as this:

for (i = 0; i < 10; i++)
{
    printf ("Hello\n")
}
printf("World");