I am new to Xcode development tool. To debug a problem, I have tried a very simple code:
int main()
{char N;
char M;
scanf("%c",&N);
scanf("%c",&M);
printf("%c",N);
printf("%c",M);
}
But the problem is that the compiler doesn't seem to read the second scanf. So I can enter one character in the console, and then the program stops. Amazingly, when I write the same code with "int" in place of "char", it works just fine. Does anybody have an idea what's wrong?
You're pressing Enter after typing a character into the first scanf, right? That Enter is what gets read by the second scanf. And printed by the second printf.
You haven't shown the output of this program, but if you change the printf formatting to %d\n from %c, then it's probably something like the following if you type "a" and press Enter:
97
10
That 10 is the character code for Enter (newline or line feed).
Related
printf("%c %c", getc(stdin), getc(stdin));
rewind(stdin); printf("\n");
printf("%c %c", getchar(), getchar());
output
s
s
s
s
I wrote the source code like that and I pressed s, enter, s, enter sequentially on windows console application using vs2017. Console shows me what I put and print output when I press the key. The result's appearance is like s, enter, enter, space, s, s, enter, enter, space, s. Did I use the function in wrong way?
Q1. I pressed s, enter sequentially, but why output is reversed?
Q2. At output's third line and sixth line, why is there exist whitespace like s not s?.
Explaining the output
The arguments to a function can be evaluated in any order that the compiler chooses.
The original code looked like this (missing printf("\n"); compared to the revised code):
printf("%c %c", getc(stdin), getc(stdin));
rewind(stdin);
printf("%c %c", getchar(), getchar());
Your 'output' in the question is extremely difficult to read and interpret. However, I'm reasonably sure that what you show consists of both your input and the program's output.
You say you type s enter twice. The output you get, copied from the question, is:
s
s
s
s
The first s and the newline that follows are what you typed. The blank line is there because the newline was read by the getc(stdin) that appears first in the argument list, then there's the blank from the format string, and the s which was read by the getc(stdin) that appears second in the argument list. Your compiler, it seems, evaluates the arguments right to left — but you should not rely on that.
Now we run into a problem; there's a newline after the second s that isn't accounted for by the code you showed when I wrote this answer. Since then, you've added the printf("\n"); which explains the output you saw.
Putting that aside, the third s is followed by a newline and is what you typed. And again, the blank line is from the newline and then space and s is the rest of the output, and there's probably a newline after that too.
JFTR, when I run your code (program getc73 compiled from getc73.c), I see:
$ getc73
s
ss
s$
$
The $ is my prompt; the first s is what I type; then the newline, blank and s are printed (with no newline at the end); then the second s on the third line is my next s and return, followed by the newline and blank s. Since there's no newline at the end, my prompt appears immediately after the fourth s. I hit return to print another prompt at the start of the line.
Why rewind(stdin) is a no-op
You can't successfully execute an fseek() call on a terminal device. They don't have any storage so you can't move around like that. The rewind() function is effectively:
void rewind(FILE *fp)
{
fseek(fp, 0, SEEK_SET);
}
So, the rewind(stdin) call fails to rewind, but it has no way to report that to you because it returns no value. It does no good; it doesn't do any harm either. If it had worked, you would not have needed to enter the information twice.
If the standard input had been a file instead of a terminal, then the rewind would succeed — disks are seekable.
Improved formatting
If you were on a POSIX system, I'd suggest you used a feature of POSIX printf() that allows you to print the same argument more than once:
#include <stdio.h>
int main(void)
{
printf("[%1$c] %1$d [%2$c] %2$d\n", getc(stdin), getc(stdin));
printf("[%1$c] %1$d [%2$c] %2$d\n", getchar(), getchar());
return 0;
}
When I run that on my terminal (it's a Mac running macOS Mojave 10.14.6 using GCC 9.2.0), I get:
s
[
] 10 [s] 115
s
[
] 10 [s] 115
The lines with just the s on them are what I typed; the other lines show that the newline is printed before the letter s.
This probably won't work for you in VS2017 — the manual for printf() formats on Windows strongly suggests it won't work. Your best bet, probably, is to use:
#include <stdio.h>
int main(void)
{
printf("%d %d\n", getc(stdin), getc(stdin));
printf("%d %d\n", getchar(), getchar());
return 0;
}
That generates:
s
10 115
s
10 115
Choosing a good format to show what you're seeing is an art form. Don't forget to include newlines at the end of (most) printf() format strings. Obviously, if you're building up a line of output piecemeal, you don't want newlines at the end of intermediate formats, but the majority of printf() statements should have a newline at the end of the format.
I know this code is not completed yet, but I cannot go any further because of this issue.
If you execute the code with any compiler you will see it.
After the instructions gets written at console, when you enter a word, loop takes 2 turns. It reduces chance 2 times too when it's supposed to be 1. Why is that?
I am using devc++ and windows.
#include<stdio.h>
#include<stdlib.h>
int main(){
int i,j,totalTrial=6,currentTrial=0;
char myWord [6]={'d','o','c','t','o','r'};
char lineArray [6]={'-','-','-','-','-','-'};
char guess;
printf("Hello,this is a simple word-guessing game. Try to find my secret word. You have 6 chances.");
printf("Lets begin!!\n");
printf("Word:\n------\n");
for(i=0;i<=6;i++)
{
printf("\nGuess a letter: ");
scanf("%c",&guess);
for(j=0;j<7;j++)
{
if(guess==myWord[j])
{
lineArray[j]=guess;
}
}
currentTrial++;
printf("\nResult: %s, %d hakkin kaldi.\n",lineArray,totalTrial-currentTrial);
}
getch();
return 0;
}
This is happening because the scanf() is reading the stray \n (newline character) from the input buffer. [When you are giving input, you must be entering a character followed by ENTER key.]
To resolve this, add a space before % character in scanf() like this:
scanf(" %c", &guess);
This will skip the leading whitespace characters (including newline character) and read the input given by the user.
With regard to the line:
scanf("%c",&guess);
How many characters do you think are turning up when you enter, for example, dENTER? I'll give you a hint, it isn't one :-)
The problem is that your scanf will read each character in turn and process it, including the newline generated when you hit ENTER.
A better solution would be to use a more complete input solution such as this one here.
It will handle many scenarios that a simple method based on scanf or getchar.
I am programming in C and i have a problem when i run a program in the cmd terminal. here is the code i use:
#include <stdio.h>
int main() {
int num;
printf("enter a number: ");
scanf("%i\n", &num);
for(int n = 1; n < num + 1; n++){
printf("%i\n", n);
}
return 0;
}
Generally, everything works like it should, exept for one thing.
when I enter a number, nothing happens. there is no output, until I write anything and press Enter, and only then the number appear.
this is a screenshot of what it looks like.
here is enter the number (and press enter) but nothing happens: http://prntscr.com/deum9a
and this is how it looks like after i entered something random nad all the numbers popped up: http://prntscr.com/deumyn
if anyone knows how to fix this, please tell me (:
Remove the \n from scanf()
scanf("%i", &num);
When you have a whitespace character in the format string, scanf() will ignore any number of whtiespaces you input and thus the ENTER you do doesn't terminate the input reading. Basically, you'll be forced to input a non whitespace character again in order complete the scanf() call.
Generally, scanf() is considered bad for input reading. So, considering using fgets() and parsing the input using sscanf().
See: Why does everyone say not to use scanf? What should I use instead?
Hi I just started learning C programming in gcc compiler on my Debian system. Here is the code
main()
{
fflush( stdin );
int a,b;
scanf("%d,%d",&a,&b);
printf("%d,%d",a,b);
}
The scanf doesn't take input for the second variable. I press 2 and then return key and it displays
root#debian:/home/wis# ./test
2
2,0root#debian:/home/wis#
I have used space and tab key also. Please help me.
You defined your scanf string as "%d,%d", so the program expect an input like 1,2.
If you give it only one digit and press Enter, it parses the first digit and leaves the second one untouched. It was assigned 0 on declaration, so that's what you are seeing when printing.
Your printf statement would benefit from an "\n" at the end, and your code snippet needs indentation. Please show your includes (#include <stdio.h>) next time, it makes it easier for us to compile and run the code.
This question already has answers here:
C: function skips user input in code
(2 answers)
Closed 8 years ago.
So for a class I am taking I have to learn C and one program I am trying to make is a simple print face program that takes 3 inputted characters and uses them to create a face.
However, whenever I run it, it asks for the eye character, then prints out the "Enter nose character: " but never takes any input, instead skipping right to the mouth character. I have looked over the code and cannot figure out what is causing this.
#include <stdio.h>
void PrintFace(char eye, char nose, char mouth) {
printf("\n %c %c\n", eye, eye); // Eyes
printf(" %c\n", nose); // Nose
printf(" %c%c%c%c%c\n",
mouth, mouth, mouth, mouth, mouth); // Mouth
return;
}
int main() {
char eyeInput;
char noseInput;
char mouthInput;
// Get character for eyes
printf("Enter eye character: ");
scanf("%c", &eyeInput);
// Get character for nose
printf("Enter nose character: ");
scanf("%c", &noseInput);
// Get character for mouth
printf("Enter mouth character: ");
scanf("%c", &mouthInput);
// Print the face using the entered characters
PrintFace(eyeInput, noseInput, mouthInput);
return 0;
}
This is the output I get:
Enter eye character: o
Enter nose character: Enter mouth character: l
o o
lllll
It seems to skip the second scan statement but I can't see why. :/
Because the input stream is line-buffered, you need to press Enter after typing in the character. Now scanf reads a single character from the stream. However, there's still a newline in the stream, and that gets picked up on the next read.
One approach is to use fgets and read a whole line of text, then pick out the first character. However, doing this properly might be a little over the top.
It might be easier if you just use code to ignore characters up until the newline, as suggested here: C code for ignoring the enter key after input. Also, you should consider using getchar or getc instead of scanf. Just make a simple function to do all this stuff, and call it whenever you want to read a character.
The carriage return you're passing by hitting "Enter" after your first character is considered a second character input. Notice the difference in carriage returns in your output.
See the linked question at: C: function skips user input in code
If I remember well, scanf does not take '\n' in a string. So you put for example "dog" as a first entry but you type an enter at the end. So the second scanf take that '\n' in the buffer shiting your program. Solution? Clean up your buffer. If you are over windows fflush() can save you, using fflush(stdin) after each scanf. Over unix fflush() does not work like that and you have to do it manually. An easy way is to put a getc() or something like that that consumes that '\n'