I'm new to programming in C and I'm trying to build and execute my first programs. My first program was Hello, World printed in the CMD like many of you. It worked great. Now it's onto bigger and better projects though and I'm having a weird issue.
I'm trying to make a basic addition calculator using standard IO operations (scanf and printf) and it won't function properly.
My program asks for 2 numbers to be input by the user and it will then display the output of said calculation. The program executes flawlessly until the scanf expression comes into play. After I input my 2 numbers to be added the CMD prompt just closes without warning and never spits out an answer or the text I have to be displayed afterward.
I tried multiple solutions to fix this problem including copying and pasting the source code directly from the website I was learning from, even their perfect code nets the same outcome..just a crash before an output is displayed. I'm posting today because I'm wondering where my issue is coming from because I'm just not sure why this program won't execute as it's supposed to. Thanks in advance and heres the code I'm working with:
#include <stdio.h>
int main()
{
int a, b, c;
printf("Enter two numbers to add\n");
scanf("%d%d", &a, &b);
c = a + b;
printf("Sum of the numbers = %d\n", c);
return 0;
}
Once console application returns from main method, the associated console window closes automatically. I assume you are using Windows OS. In that case add a system("pause"); before your return 0; statement.
For platform independent solution you can just show a prompt to user and wait for a key press before returning from main. As #chux pointed out in comment any character remaining in input buffer (enter from scanf in this case) must be cleared.
#include <stdio.h>
int main()
{
int a, b, c;
printf("Enter two numbers to add\n");
scanf("%d%d", &a, &b);
c = a + b;
printf("Sum of the numbers = %d\n", c);
//clear input buffer
int d;
while ((d = getchar()) != '\n' && d != EOF) { }
printf("Press ENTER key to Continue\n");
getchar();
return 0;
}
CMD prompt just closes without warning and never spits out an answer
Seems like you are opening a new terminal on Windows machine. I compiled the code and it works. Your program closes just after printing the answer so you simply cannot see it. Stop it artificially before the end. To prove it add the following line directly before return:
scanf("%c", &a);
This will result in stopping the program and waiting for input. You will need to enter another number which will essentially be ignored but will stop the program so you can see the output. This is not a good target solution though due to you need to enter some characters to go on and exit but it proves the point :)
Related
I have just installed visual studio code & after watching some basics, I have written a simple C code to calculate area of rectangle. But, when I run the code its taking too long to execute.
My system configuration are 4gb RAM, i3 - 5th gen (poor boi) Here is the code:
#include <stdio.h>
int main()
{
int l;
int b;
printf("enter the length of rectangle in an integer value");
scanf("%d ", &l);
printf("enter the breadth of rectangle in an integer value");
scanf("%d ", &b);
printf("the area of rectangle is %d", l * b);
return 0;
}
Change scanf("%d ", &l); to scanf("%d", &l); and scanf("%d ", &b); to scanf("%d", &b);.
A space in a format string tells scanf to read from input until a non-white-space character is read. So, after scanf("%d ", &l);, the program continues reading input until you type something such as the next number. If you do type that next number, then scanf("%d ", &b); reads it and then continues reading until it sees another character that is not a space. As long as you do not type anything, or type only spaces and returns/enters and other keys that generate white space characters, the program continues waiting.
Removing the spaces eliminates this.
The program is not “taking too long to execute.” In fact, the program is waiting for user input, so it is just waiting, not executing. Because it did not complete, you concluded it was taking too long to execute. Instead of reporting a conclusion, you should have reported the observed behavior: After typing input, there was no visible activity by the program.
Skipping spaces before performing a conversion is built into most scanf conversions, such as %d. Only include a space character in a format string where you want to skip spaces that would not be skipped normally, such as before a %c conversion, which does not have the automatic skipping built in.
Also, in printf("the area of rectangle is %d", l * b);, add a new-line character at the end, printf("the area of rectangle is %d\n", l * b);. (This ensures the prompt printed by a command-line shell after the program finishes executing is on a new-line, so it will not be confused with the program output. C is designed to end lines of output with new-line characters, so you should make it a regular practice.)
Alright guys, I found the solution. The first thing I want to apologize is that the code is not taking too long to execute. The code is actually executing but there was no visible activity. Everything in the code is just perfect. The mistake I made is I didn't change one setting in VS Code. I use code runner extension to run my codes [that's the most popular]. I have to go in File --> Preferences --> Settings --> Type 'code runner' in the search bar --> check the setting given in this image - https://drive.google.com/file/d/1QwB2NjKpYX7M0b5w55sqy4oMVAOk_IK2/view?usp=sharing
Lot of people were trying to figure out a mistake in the code, but there is nothing wrong with it. Code runner was not able to run in the terminal, so there was no activity to take any user input - due to read only text editor. After changing that setting I was able to give user input directly in the terminal.
Credit for the solution - https://www.youtube.com/watch?v=Si8rN5J249M
EDIT: There is a mistake in my code also lol, which is mentioned in another answer - Really appreciate it Eric Postpischil
#include<stdio.h>
#include<conio.h>
int main()
{
clrscr();
int a,b,c;
printf("enter the 2 numbers: ");
scanf("%d %d",&a,&b);
c=a+b;
printf("the sum is : %d ",c);
return(0);
}
this is a simple program to add 2 numbers.
my program would let me input the value..but it would not print the sum ,nor would it print the next line.
it would run till the scanf() and as i press enter, it would jst exit the program.
can you please tell me whats wrong. I am a beginner programmer...
There are two things you should think of here.
End printouts with a newline character, because stdout is often line buffered. Do printf("the sum is : %d \n",c); instead. Or call fflush(stdout); expliticly after the printout. This will ensure everything gets printed.
Add some input code in the end. Like an extra scanf("%d", &a); This is basically a small hack to prevent the window from closing before you can see the final output. Another alternative is to add sleep(3); to sleep for 3 seconds. A third alternative here is to see if there are some settings that controls the closing of the window in your IDE.
Your program works correctly, but it exits right after printing the output, giving you no time to look at it.
Consider adding some input before return(0);, such as 2 getchar(); calls. You need 2, because the first character read will be the \n that you typed after the numbers.
Just started learning C and I have a problem with the scanf function. Every time I enter a number in the console, it will be printed right under the input. The program still works, but it is a little bit annoying.
(I am using CLion from JetBrains)
int main()
{
int x;
printf("Number: ");
scanf("%d", &x);
printf("Your number is %d!", x);
}
This is the output:
Number:15
15
Your number is 15!
Process finished with exit code 0
It is an issue in clion (Why is CLion printing back inputs from standard input?). Currently unresolved. This problem exist for C and C++.
This bug resides for four years. I definitely advice you to change your compiler if you are not bound this for a particular reason.
I'm currently in the very early stages of learning C programming and am working my way through "Beginning Programming with C for Dummies" using Code::Blocks.
The first activity in Chapter 7, Fetching characters with getchar(), asks us to copy the code exactly as it's presented in the book; see below:
#include <stdio.h>
int main()
{
int c;
printf("I'm waiting for a character: ");
c = getchar();
printf("I waited for the '%c' character.\n", c);
return (0);
}
The output I get is:
I'm waiting for a character:
However according to the book, the output that I should be seeing is the character's ASCII code value. It then asks that I change the %c placeholder to %d to display the value, but still I get the same outcome as before. I could probably recite the code with my eyes closed I've checked it through that may times; I simply cannot see where I'm going wrong.
Am I right in thinking that the getchar() function isn't being recognized? Or that the code isn't being read after the first printf statement? Any guidance is welcome as I don't want to move on until I've understood the problem.
Please enter any key, then 2nd printf will show the result. getchar() is expecting input from user, controller reach at 2nd line & waiting for input.
I am using Code::Blocks for programming and yeah i am a beginner but everytime i write a program it pauses in IDE but does not pause while executing directly.
What is the possible reason ? Can anyone explain me ?
My code goes as follows :
#include <stdio.h>
void main()
{
float length,breadth,Area;
printf("Enter the value of length and breadth \n\n");
scanf("%f %f",&length,&breadth);
printf("You entered length=%f and breadth=%f \n\n",length,breadth);
Area= length * breadth;
printf("The area of the rectangle is %f\n\n",Area);
return 0;
}
You have to tell your program to wait for input at the end otherwise it will execute, do exactly what you wrote in your code and exit. The "good" way would be to execute it from a terminal (cmd if you are on windows)
#include <stdio.h>
int main()
{
float length,breadth,Area;
printf("Enter the value of length and breadth \n\n");
scanf("%f %f",&length,&breadth);
getchar(); // catch the \n from stdin
printf("You entered length=%f and breadth=%f \n\n",length,breadth);
Area= length * breadth;
printf("The area of the rectangle is %f\n\n",Area);
getchar(); // wait for a key
return 0;
}
Why do you need a getchar() after your scanf()?
When you enter your numbers you finish it with a press of enter. Let's see what you are reading: a float whitespaces and another float. The \n is not consumed by scanf(), but left in the input buffer (stdin). The next time you use a function that reads from stdin, the first sign this function sees is a \n (Enter). To remove this \n from the input buffer you have to call getchar() after scanf() which reads 1 character from the input buffer. I'm sure you will encounter this behaviour more often in future.
As a really, REALLY bad practise, you can just use a getch call to stop the excecution (or any function that generates a small pause, getch is not a standard function).
A program is not supposed to pause after execution and this is a feature added by the IDE. If you want execution to pause and wait for some input you should instruct it to do so. For instance if you are on windows you can add a line:
system("pause");
Right before return 0;. This is not advisable, but may help you for debugging in some cases. Also the standard requires that your main function is int, not void. So you better get used to writing int main instead of void main.
You need to use getch(); at the end of program (Before return statement)
Getch holds down the output screen.