Why this program can't working?

Question

1.00/5 (1 vote)

See more:

C

#include <stdio.h>

int main()
{
    int num;
    float input[3],sum;
    printf("Please enter a limit number :");
    scanf("%d",&num);

    for(int i=1;i<=num;i++){
        for(int j=0;j<=2;j++){
            scanf("%1.1f",&input[j]);
            sum=(input[0]*2.0)+(input[1]*3.0)+(input[2]*5.0);
        }
        printf("%1.1f\n",sum/10.0);
    }
    return 0;
}

What I have tried:

I want to find weighted average but i can't give input on this program when i use scanf("%1.1f",&input[j]);(in bold and underline). Now what to do and why??

Posted 1-Apr-22 7:02am

Fardin Khan_451

Updated 1-Apr-22 7:20am

k5054

v3

Add a Solution

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

k5054 · Accepted Answer · 2022-04-01T07:16:00

The scanf() format specifier "%1.1f" is invalid. For scanf you can specify a field width, but not a precision (i.e. digits after the decimal). Even if you could specify precision, the format "%1.1f" is problematic: it defines a field that is 1 character wide and has 1 digit after the decimal. The smallest input string that one could use would be ".1", which is at least 2 chars wide, so its not clear what scanf would do to satisfy both the field width of 1 char and 1 char after the decimal.

Patrice T · Accepted Answer · 2022-04-01T07:20:00

Quote:
but i can't give input on this program when i use scanf("%1.1f",&input[j]);(in bold and underline). Now what to do and why??

You may have to read documentation about scanf.
Try

C++

scanf("%1.1f",&input[j]);

Another problem

C++

for(int i=1;i<=num;i++){
    for(int j=0;j<=2;j++){
        scanf("%1.1f",&input[j]);
        sum=(input[0]*2.0)+(input[1]*3.0)+(input[2]*5.0); // this line
    }
    sum=(input[0]*2.0)+(input[1]*3.0)+(input[2]*5.0); // goes here
    printf("%1.1f\n",sum/10.0);
}

Why this program can't working?

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0