Answer / janava

In this a variable is a integer but read the int as char so
it shows error.
this can be written as
#include<stdio.h>
#include<conio.h>
void main()
{
int a;
printf("\n Enter a number");
scanf("%d",&a);
getch();
}

Answer / lalithpriya

#include<stdio.h>
#include<conio.h>
void main()
{
int a;
clrscr();
printf("Enter a number:");
scanf("%d",&a);
printf("The entered number is %d",a);
getch();
}

Answer / r.kumar bsc

getting a value is defined %d that is scanf("%d\n");
this formatted is correct that %c is defined only is
getting the characters.

Answer / aarti

#include<stdio.h>
#include<conio.h>
void main
{
clrscr();
int number;
printf("\n enter a number:");
scanf("%d",&number);
getch();
}

Answer / pawanjha12

Here in the scanf("%c\n");
it is just take a input. and, it will store that input
value anywhere.
it would be just garbage value. ok.........
nothing is output.
say, null pointer assignment.........

Answer / gaurav tyagi

there is a error in program because ""is use in headerfile
due to this compiler is search the conio.h&stdio.h file in current directory
note:<>is use to tell compiler find these file in system area

Answer / aarti

#include<stdio.h>
#include<conio.h>
void main()
{
int number;
clrscr();
printf("\n enter a number:");
scanf("%d",&number);
getch();
}

Answer / mohit

#include {studio h}
#include{conio h}
[void main]

Answer / sansiri

it access the value but it doesnot assign for any variable

Answer / lodu

#INCLUDE<STDIO.H>
#INCLUDE<CONIO.H>
{
VOID MAIN()
{
INT X,Y,Z;

Declaration of Cube Guys please help me.. Is this a right way to declare cube.? If i Compile it. It Says: Cube undeclared what should i do? Please help \thanks in advanced #include<stdio.h> #include<math.h> #include<conio.h> main( ) { float x,y; while(x++<10.0) { printf("Enter Number:"); scanf("%d", &x); y = cube(x); printf("%f %f %f \n", x,pow(x,2),y); cube(x); } { float x; float y; y = x*x*x; } getch(); return (y); }

2 Answers  

I am using Qt 5.6 during compilation it stops and gives error about Qmake The process "C:QtQt5.6.35.6.3msvc2015_64inqmake.exe" crashed. Error while building/deploying project untitled1 (kit: Desktop Qt 5.6.3 MSVC2015 64bit) When executing step "qmake"

0 Answers  

Assume that the int variables i and j have been declared, and that n has been declared and initialized. Write code that causes a "triangle" of asterisks of size n to be output to the screen. Specifically, n lines should be printed out, the first consisting of a single asterisk, the second consisting of two asterisks, the third consistings of three, etc. The last line should consist of n asterisks. Thus, for example, if n has value 3, the output of your code should be * ** *** You should not output any space characters. Hint: Use a for loop nested inside another for loop.

2 Answers   HCL,

what is the error in the following code: main() { int i=400,j; j=(i*i)/i; }

4 Answers  

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }

19 Answers  

what is the large sustained error signal that eventually cause the controller output to drive to its limit

1 Answers   TCS,

#include<iostream.h> #include<stdlib.h> static int n=0; class account { int age,accno; float amt; char name[20]; public: friend void accinfo(account [] ,int); void create(); void balenq(); void deposite(); void withdrawal(); void transaction(account []); }; void account :: create() { static int acc=1231; accno=acc+n; cout<<"\n\tENTER THE CUSTOMER NAME : "; cin>>name; cout<<"\n\t ENTER THE AGE : "; cin>>age; cout<<"\n\t ENTER THE AMOUNT : "; cin>>amt; // if(amt<=500) // cout<<"\n\tAMOUNT IS NOT SUFFICIENT TO CREATE AN ACCOUNT..."; cout<<"\n\t YOUR ACCOUNT NUMBER : "<<accno<<endl; n++; } void accinfo(account cus[],int ch) { int no,flag=0; cout<<"\n\t\tENTER YOUR ACCOUNT NUMBER : "; cin>>no; for(int i=0;i<=n&&flag==0;i++) if(no==cus[i].accno) { flag=1; switch(ch) { case 2: cus[i].balenq(); break; case 3: cus[i].deposite(); break; case 4: cus[i].withdrawal(); break; case 5: cus[i].transaction(cus); break; default: cout<<"\n\t\tEND OF THE OPERATION"; exit(1); } } if(flag==0) cout<<"\n\t\tYOUR ACCOUNT DOES NOT EXIST..."<<endl; } void account :: balenq() { cout<<"\n\t\tCUSTOMER NAME : "<< name << endl; cout<<"\n\t\tBALANCE : "<< amt << endl; } void account :: deposite() { int damt; cout<<"\n\t\tCUSTOMER NAME : "<< name <<endl; cout<<"\n\t\tBALANCE : "<< amt <<endl; cout<<"\n\tENTER THE AMOUNT TO BE DEPOSITED : "; cin>>damt; amt+=damt; cout<<"\n\t\tYOUR CURRENT BALANCE : "<<amt<<endl; } void account :: withdrawal() { int wamt; cout<<"\n\t\tCUSTOMER NAME : "<< name; cout<<"\n\t\tBALANCE : "<< amt; cout<<"\n\tENTER THE AMOUNT TO BE WITHDRAWN : "; cin>>wamt; if(amt-wamt>=500) { amt-=wamt; cout<<"\n\t\tYOUR CURRENT BALANCE : "<<amt; } else cout<<"\n\tYOUR BALANCE IS TOO LOW FOR WITHDRAWAL..."<<endl; } void account :: transaction (account cus[]) { int no,tamt,flag=0; cout<<"\n\tENTER THE RECEIVER'S ACCOUNT NUMBER : "; cin>>no; cout<<"\n\t\t ENTER THE AMOUNT : "; cin>>tamt; for(int i=0;i<=n&&flag==0;i++) if(cus[i].accno==no) { flag=1; cus[i].amt+=tamt; amt-=tamt; cout<<"\n\t\tYOUR CURRENT BALANCE : "<<amt<<endl; cout<<"\n\t\t RECEIVER'S BALANCE : "<<cus[i].amt<<endl; } if(flag==0) cout<<"\n\tRECEIVER'S ACCOUNT NUMBER IS NOT AVALIABLE..."<<endl; } void main() { account cus[10]; int ch; do { cout<<"\n\t\t BANK ACCOUNT"; cout<<"\n\t\t ************\n"; cout<<"\n\t\t1.CREATE AN ACCOUNT"; cout<<"\n\t\t2.BALANCE ENQUIRY"; cout<<"\n\t\t3.DEPOSITE"; cout<<"\n\t\t4.WITHDRAWAL"; cout<<"\n\t\t5.TRANSACTION"; cout<<"\n\t\t6.EXIT\n\n"; cout<<"\n\t\tENTER YOUR CHOICE : "; cin>>ch; if(ch==1) cus[n].create(); else accinfo(cus,ch); }while(1); }

1 Answers  

What is the code for following o/p * * * * * * * * * * * * * * * *

1 Answers  

Find the error (2.5*2=5) (a) X=y=z=0.5,2.0-5.75 (b) s=15;

3 Answers  

loop1: { x=i<n?(i++):0; printf("%d",i); exit(x); continue; } Error- misplaced continue. Doubt-1.will the exit(x) be executed for all values of x 2.will this statement go out of the program.

5 Answers   CMC,

void main() { int i=5,y=3,z=2,ans; clrscr(); printf("%d",++i + --z + i++ + --i * ++y); i=5,y=3,z=2; ans=++i + --z + i++ + --i * ++y; printf("\n%d",ans); getch(); } Its output is 37 and 31.... Please explain me why its different How it works.....

2 Answers  

errors are known as?

3 Answers   EX, State Bank Of India SBI,