The code is:
int main() {
scanf("%d");
puts("hello!");
getch();
return 0;
}
If I input a number like 7, the program crashes but if I input a character like h, the program executes successfully.
Could you please explain me the reason behind it?
I am using MinGW on Windows XP.
Access specified %d won't get other character except number.
If you give other character the scanf get executed and nothing will store.
When you give numeric character the scanf get executed and try to store the value.
Due to storage location is unavailable the program get crashed.
Try to use the storage location in the scanf like
scanf("%d",&a);
The every version say warning to storage location.
So try to check warnings.
Compile with the warnings.
Read the manual page you will not that scanf("%d"); will require an address to put the number into if there is one
Try
int x;
scanf("%d", &x);
You should also check the return value from scanf as per the manual page. I leave that as an exercise to the reader to do with as he will.
Try
check = scanf("%c %d",&a, &b);
printf("check = %d \n",check);
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
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 :)
I'm writing a program in C that displays a menu that asks the user to make a decision.
I want to bring up the menu multiple times in the program so I put it into function.
Each choice has a number associated with it:
(1) Add
(2) Subtract
The function displays the menu and then then I scanf the int response to a local variable and then return the variable (I of course declare the variable at the beginning of the function).
int function ()
{
int choice;
// *insert print menu code here*
scanf(" %i", &choice);
return choice;
}
Can I make this shorter by somehow doing a: return scanf(" %i", stdin);
I'm not sure if stdin would be the right choice, but that's what searching tells me.
Sure you can:
int scanInt(void) {
int result;
return (scanf(" %i", &result), result);
}
But is it useful? Probably not.
Note that this might look like undefined behaviour at first glance, because you're writing and reading from the same memory location in one expression. But it is not, as the comma operator introduces a sequence point.
You can't do this with scanf: you'd be returning the return value of scanf, which only indicates the number of items successfully read. If you need to return choice then you'll need to call scanf to fill choice and return it in two separate steps.
Another point here is that you should check the return value of scanf: it will fail if it's unable to convert the input. In your code, scanf will fail if the input is not an integer.
Furthermore, scanf reads from stdin - standard input - by default, you can't pass that as an argument unless you're using fscanf, which otherwise behaves in the same way.
Have you looked at using fgets for your input? It does return the next input line, which would allow you to do the read in the return statement since fgets returns the next line of input as a string. But you'd still need to parse the input yourself, which should probably happen within your function. In any case, fgets is typically a better choice for user input because it separates reading the input from parsing it, and avoids problems when using scanf due to input that doesn't match your format string.
You can, sort of -- but don't.
scanf returns the number of items scanned. You can't make it return the value of one of the scanned items.
return scanf(" %i", &choice), choice;
This uses the comma operator (not to be confused with the comma delimiter between function arguments), which evaluates both operands and yields the result of the right operand.
But there is no good reason to do this. There is no great virtue in making your source code more compact like this. The multi-line form is clearer, and it's easier to modify by adding error handling (what happens if the user enters something other than a number?)
Or you could write your own function that calls scanf and returns the value you want, but that still doesn't address the issue of error handling.
And if your goal is simply to put them on one line:
scanf(" %i", &choice); return choice;
But again, this is not an improvement.
No, you can't with scanf. The function needs a memory location to store the result, and doesn't return it. So you just have to live with the extra code.
#include <stdio.h>
#include <math.h>
int main (void)
{
float inches;
printf("Enter the number of inches\n");
scanf("%f\n",&inches);
float feet;
float cm;
float yards;
float meter;
feet = 12 * inches;
cm = 2.54 * inches;
yards = 36 * inches;
meter = 39.37 * inches;
printf("Amount in feet: %f\n", &feet);
printf("Amount in cm: %f\n", &cm);
printf("Amount in yards: %f\n", &yards);
printf("Amount in meters: %f\n", &meter);
getchar();
return 0;
}
I'm using Dev c++
Is the problem i'm problem I'm working on in C. Basically enter in a number in inches then print amount in cm,yards,meters and feet. This is giving me 0.0000 or something for all of them or actually the time it is up. I can't keep the screen up and I thought that was the purpose of getchar() but I must have been mistaken. Any help is great. Thanks!
EDIT 1
What about as far as keeping dev c++ on the screen instead of closing out after I put stuff in? I am also having to put 2 values in before it returns in anything when the screen pops up? Why??
Two problems:
The usual problem with using scanf(), in that it leaves the newline after the number unread and the following read operation (the getchar() here) reads it.
You shouldn't pass pointers to printf(), but the actual values.
You're trying to print the addresses of your floats as floats, you just want to say this:
printf("Amount in feet: %f\n", feet);
Note the lack of an address (&) operator on feet. You want to apply similar changes to your other printf calls.
With printf, you don't give it the address of the float values, you just give it the values. Remove the &s from the printf calls.
You need the address in scanf because the function modifies the variables that you pass in, but printf just needs the values. As it is, the printf is essentially reinterpreting the pointers as floats, which is why you get the garbage values displayed.
About I can't keep the screen up, it's a common problem to everyone trying to execute a console program in a graphical environment directly from the IDE, in particular Dev-C++. The problem is that there's no console for I/O, then one is provided but just for the time the program is running, and since programs are fast, if you do not add a pause after your last input and output, you won't have the time to read the output.
Many MS Windows Dev-C++ users add an horrorific system("pause"). I always suggest that, if Dev-C++ is not able to provide a console for I/O with the option "keep it opened even after the program ends", then it is better you open a shell (cmd or powershell on windows) and run your program directly from there.
About the input problem, unluckly scanf-ing has several buffering problem since the input that is not recognized for the given format is not discarded and is ready for the next reading. E.g.
scanf("%f", &aFloat);
scanf("%f", &theNextFloat); // in your case you have the extra getchar();
won't stop for the second scanf if you write 1.25 4.5 as first input, since the 4.5 is already available for the next scanf. In your case it was a newline that was left in the buffer and since getchar has found it, it does not need to wait for input. You could use a
while( getchar() != EOF ) ; instead, and then to exit you need to hit Ctrl-D.
I created a very simple progam whith a menu,
that take a value, then memorize it into the
local variable value, and finally with the
second option the progam prints the value.
my question is:
Why does the program work only if I add an "h"
to the scanf parameter?
In other words: what kind of relation there is
between scanf() and my local int value variable?
thanks!
p.S. (I used Dev-C++ (GCC) to compile it.
With Visual Studio it works)
#include <stdio.h>
main () {
int value = 0;
short choice = 0;
do {
printf("\nYour Choice ---> ");
scanf("%d", &choice); /* replace with "%hd" and it works */
switch (choice) {
case 1:
printf("\nEnter a volue to store ");
scanf("%d", &value);
getchar();
printf("\nValue: %d", value);
break;
case 2:
printf("\nValue: %d", value);
break;
}
} while (choice < 3);
getchar();
}
With scanf, the "h" modifier indicates that it's reading a short integer, which your variable choice just happens to be. So the "%hd" is necessary to write only two bytes (on most machines) instead of the 4 bytes that "%d" writes.
For more info, see this reference page on scanf
The variable choice is of type short so that's why you need the %h specifier in scanf to read into it (in fact you don't need the d here). The int type just requires %d. See the notes on conversions here
You're reading into a short. The h is necessary because %d is the size of an int by default. See this reference page on scanf.
It looks like your problem is that choice is a short, which is (generally) 2 bytes long, while %d expects an integer, which is (generally) 4 bytes long… So the scanf clobbers whatever comes after choice on the stack.
choice is a short and %d specifies an int.
When you specify %d, scanf has to assume that the associated argument is a pointer to an int sized block of memory, and will write an int to it. When that happens it will likely be writing to data adjacent to but not part of choice and the results are undefined and probably not good! If it works in one compiler and not another that is simply the nature of undefined behaviour!
In GCC -Wformat should give you a warning when you make this error.
From the comp.lang.c FAQ:
Why doesn't the code short int s; scanf("%d", &s); work?
Someone told me it was wrong to use %lf with printf. How can printf use %f for type double, if scanf requires %lf?
%d is for reading an int, not a short. Your code never really "worked" -- it just appears that in this case you didn't notice any difference between what you wanted and the undefined behavior you got.
The modifier for scanf to input a variable of type short is %hd. Hence you need to specify the correct modifier.
scanf("%d",&integer); // For integer type
scanf("%hd",&short_int); // For short type
Hence it doesnt work.
Depending upon numeric padding, endian-ness, and other such issues, you may be storing either the upper or lower part of the input value into choice; you are storing the rest of the input value into memory that may or may not be being used for anything else.