C program in ubuntu using visual studio code

Question

The code I wrote:

 #include<stdio.h>

 int main()
 {
 int i,m,maths=0,english=0,e,science=0,s,all=0;
 for(i=1;i<=5;i++)
 {
    printf("enter the marks of english:\n");
    scanf("%d",&e);
    printf("enter the marks of maths:\n");
    scanf("%d",&m);
    printf("enter the  marks of science:\n");
    scanf("%d",&s);
    if(m>=95 && s>=95 && e>=95)
     all++;
    if(m>=90)
    maths++;
    if(e>=90)
    english++;
    if(s>=90)
    science++;

 }
  printf("95% or above in all subject : %d \n",all);
   printf("90% or above in maths :%d\n",maths);
  printf("90% or above in english: %d\n",english);
  printf("90% or above in science: %d\n",science);
 }

I have to input the marks of 5 students through keyboard and I am getting the following error that I don't understand. Why am I getting this kind of error?

 gcc studentMarks.c -o studentMarks.out 
studentMarks.c: In function ‘main’:
   studentMarks.c:24:17: warning: ' ' flag used with ‘%o’ gnu_printf format [-Wformat=]
   printf("95% or above in all subject : %d \n",all);
             ^
  studentMarks.c:24:44: warning: format ‘%d’ expects a matching ‘int’ argument [-Wformat=]
   printf("95% or above in all subject : %d \n",all);
                                       ~^
 studentMarks.c:25:17: warning: ' ' flag used with ‘%o’ gnu_printf format [-Wformat=]
 printf("90% or above in maths :%d\n",maths);
             ^
 studentMarks.c:25:37: warning: format ‘%d’ expects a matching ‘int’ argument [-Wformat=]
 printf("90% or above in maths :%d\n",maths);
                                ~^
 studentMarks.c:26:17: warning: ' ' flag used with ‘%o’ gnu_printf format [-Wformat=]
 printf("90% or above in english: %d\n",english);
             ^
studentMarks.c:26:39: warning: format ‘%d’ expects a matching ‘int’ argument [-Wformat=]
 printf("90% or above in english: %d\n",english);
                                  ~^
 studentMarks.c:27:17: warning: ' ' flag used with ‘%o’ gnu_printf format [-Wformat=]
 printf("90% or above in science: %d\n",science);
             ^
 studentMarks.c:27:39: warning: format ‘%d’ expects a matching ‘int’ argument [-Wformat=]
 printf("90% or above in science: %d\n",science);
                                  ~^

enter image description here

Gopal Srivastava · Accepted Answer · 2020-11-19 12:04:48Z

1

In C programming %o represent octal values. So when you are writing 95% or its reading it as %o. Since there is a space between % and o, C compiler doesn't recognize it as octal or percentage and that's why its giving warnings. But, if you replace 95% with 95%% then c compiler reads it as 95 percent.

#include<stdio.h>

 int main()
 {
 int i,m,maths=0,english=0,e,science=0,s,all=0;
 for(i=1;i<=5;i++)
 {
    printf("enter the marks of english:\n");
    scanf("%d",&e);
    printf("enter the marks of maths:\n");
    scanf("%d",&m);
    printf("enter the  marks of science:\n");
    scanf("%d",&s);
    if(m>=95 && s>=95 && e>=95)
     all++;
    if(m>=90)
    maths++;
    if(e>=90)
    english++;
    if(s>=90)
    science++;

 }
  printf("95%% or above in all subject : %d \n",all);
   printf("90%% or above in maths :%d\n",maths);
  printf("90%% or above in english: %d\n",english);
  printf("90%% or above in science: %d\n",science);
 }

This should work now.

answered Nov 19, 2020 at 12:04

Gopal Srivastava

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1 Comment

apeksha shakya Over a year ago

yes it worked .Thankew @Gopal_Srivastava

Jeff Holt · Accepted Answer · 2020-11-19 12:39:35Z

0

For each line, from lines 24 through 27, the compiler is emitting two warnings (not errors as you indicate) to help you out.

The first is caused when your compiler encounters % o from the longer substring 95% or.

The compiler and the printf function interprets % o as %o with a blank flag to allow for a - to be emitted if the signed numeric type is negative.

The second warning is caused because of the first interpretation and the fact that you included only one argument after the format string. If you had accidentally provided two signed integer arguments after the format string, then you would not have seen any warnings.

If you try to execute the code as is in spite of the warnings, then you'd get undefined behavior.

Any time you want to include a % in your format string and you want to emit it literally, then you have to encode it as %%.

As an aside, unrelated to your question, the value you provided to the gcc -o option is unusual. One would typically call the output executable studentMarks or studentmarks, not studentMarks.out.

edited Nov 19, 2020 at 12:39

answered Nov 19, 2020 at 12:32

Jeff Holt

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1 Comment

apeksha shakya Over a year ago

yes you are right , it worked when i used %% .

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