Answer / pavan_mustyala

1. Syntax Errors: Error found during compilation e.g.
missing semi colon, missing comma, undeclared identifier,
etc

2. Semantic Errors: Logical error in the program due to
which functionality behaves differently than expected.

3. Runtime Errors: Exceptions due to insufficient memory
and Segmentation fault, stack over flow, etc.

Answer / sivavendra

compilation errors(occurs during compilation) and runtime errors(occurs during runtime)...

Answer / vishwjeet singh

PROTOTYPE ERROR:when any header file is missing then we error hat was shown by compilling the program is prototype error.
SYNTEX ERROR: Actually these type of errors occured when
programmer misses the syntax.(Syntsx: (,),(.),(;),("") etc.)
Example:- printf("your sum is %d"c);
This may occure error "Syntax error".
Right Syntax is:printf("your sum is %d",c);

SYSTEMATIC ERROR:These type of errors occure when there is a
missmatch of functionallity.
Example: int sum is;
this may occure error because while declaring a variable c
is not allowed space( ).
Right Syntax is: int sum_is;

RUNTIME ERROR: Thes etype of errors also known as "linker
errors".These type of errors occure after compilation.This
will not give you the line where these type of errors occured.

Answer / sudhir seth

Syntax errors: Actually these type of errors occured when
programmer misses the syntax.(Syntsx: (,),(.),(;),("") etc.)
Example:- printf("your sum is %d"c);
This may occure error "Syntax error".
Right Syntax is:printf("your sum is %d",c);

Symmentic errors:These type of errors occure when there is a
missmatch of functionallity.
Example: int sum is;
this may occure error because while declaring a variable c
is not allowed space( ).
Right Syntax is: int sum_is;

Runtime errors: Thes etype of errors also known as "linker
errors".These type of errors occure after compilation.This
will not give you the line where these type of errors occured.

To generate the series 1+3+5+7+... using C program

18 Answers  

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Using string functions write a program that will accept the name of the capital as input value and will display the corresponding country. ------------------------ Capitals Countries ------------------------ Capitals Countries Ottawa Canada Moscow Russia Rome Italy I can't not get it to run properly

1 Answers   AMA,

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I'm having trouble with coming up with the correct code. Thank You!! The assignment was to write a program using string functions that accepts a price of an item and displays its coded value. The base of the keys: X C O M P U T E R S 0 1 2 3 4 5 6 7 8 9 Sample I/O Dialogue: Enter Price: 489.50 Coded Value: PRS.UX

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WHAT WILL BE THE OUTPUT OF THE FOLLOWING QUESTION void main() { int x=4,y=3,z; z=x-- -y; printf("%d%d%d",x,y,z); }

25 Answers   HCL,

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Given an int variable n that has already been declared and initialized to a positive value, and another int variable j that has already been declared, use a do...while loop to print a single line consisting of n asterisks. Thus if n contains 5, five asterisks will be printed. Use no variables other than n and j .

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Write a program to accept two strings of Odd lengths. Then take all odd characters from one string and even characters from the other and concatenate and produce a string.

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write a profram for selection sort whats the error in it?

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#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  

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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.....

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2. A student studying Information Technology at Polytechnic of Namibia is examined by coursework and written examination. Both components of assessment carry a maximum of 50 marks. The following rules are used by examiners in order to pass or fail students. a. A student must score a total of 40% or more in order to pass (total = coursework marks + examination marks) b. A total mark of 39% is moderated to 40% c. Each component must be passed with a minimum mark of 20/50. If a student scores a total of 40% or more but does not achieve the minimum mark in either component he/she is given a technical fail of 39% (this mark is not moderated to 40%) d. Grades are awarded on marks that fall into the following categories. Mark 100-70 69-60 59-50 49-40 39-0 Grade A B C D E Write a program to input the marks for both components (coursework marks out of 50 and examination marks out of 50), out put the final mark and grade after any moderation. [30]

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what is meant for variable not found?

3 Answers  

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I can not get my C++ program to work right. It is supposed to tell if a word is a palindrome or not, but it only tells thet the word is not a palindrome. And I can't fix it.

1 Answers  

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