I have tried following code there is require %d before entering character. That is an after switch loop in code.
#include<stdio.h>
#include<conio.h>
void sum();
void mul();
void main()
{
char ch;
int c;
clrscr();
do
{
printf("\n\n Enetr choice ");
printf("\n\n\t 1: SUM \n\n\t 2: MUL");
scanf("\n\n\t %d",&c);
switch(c)
{
case 1:
sum();
break;
case 2:
mul();
break;
default:
printf("\n\n hhhh..... ");
}
printf("\n\n Want u calcualte again");
//scanf("%d");
scanf("%c",&ch);
printf("\n ch value is %c",ch);
}while(ch=='y'|| ch=='Y');
getch();
}
void sum()
{
int s;
s=10+50;
printf(" SUM: %d",s);
}
void mul()
{
int s;
s=10*50;
printf(" SUM: %d",s);
}
Here in this code after switch I tried to input character but without the scanf statement which is in comment is require while you input character. without that scanf statement compiler does not take character input. so please give me solution.
Its because you have to "eat up" the newline from previous input
You don't have to use %d.
Instead use:
while((c = getchar()) != '\n' && c != EOF) ;
in place of
//scanf("%d");
to discard the newline.
this is the problem occurred due to the insertion of next line character i.e. '\n' instead of following statement
scanf("%c",&ch);
you should use
scanf("\n%c",&ch);
Now what will happen firstly control goes to new line and then it will insert or input the character, just change this statement you will find your program execute properly...
You have to consume new line character.
You can add space before %c in your scanf statement to ignore white space
You should change
scanf("%c",&ch);
to
scanf(" %c",&ch);//this makes scanf ignore white spaces like new line, space etc.
or use getchar() to do it.
c=getchar();
For more insight go to question:
scanf() function doesn't work?
Another method that tells scanf to consume or recognize white space (and a new line is considered white space) is to code:
char ch[2];
...
scanf("%1s", &ch);
...
if (ch[0] == 'x' etc.
Related
I am trying to make a menu with the options to compress a text inputted by the user and then store that value to be extracted in the extract menu option.
The issue lies in that it seems like the code isn't following the void statements, for example
case 1: compress();//compress statement
It seems to only get the printf statement in the void compress(void) and not the scanf, which it then follows with the loop of the menu.
Any solutions?
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
void menu(void);
void compress(void);
void extract(void);
int main(void)
{
menu();
return 0;
}
void menu(void)
{
int choice;
do
{
printf("Menu\n\n");
printf("1. Compress Text\n");
printf("2. Extract Text\n");
printf("3. Exit\n");
scanf_s("%d", &choice);
switch (choice)
{
case 1: compress();//compress statement
break;
case 2: extract();//extract statement
break;
case 3: printf("Ciao\n");
exit(0);
break;
default: printf("Invalid Entry. Try again\n");
break;
}
} while (choice != 3);
}
void compress(void) {
printf("\n-------------------------\n");
printf(" COMPRESS ");
printf("\n-------------------------\n");
printf("\nPlease enter a word/sentence to be compressed:\n");
char txt[200];
scanf_s("%c", &txt);
printf("\nYour word/sentence is %c", txt, "\n");
char comp = strlen(txt);
int mask[200]{};
for (int i = 0; i < comp; ++i) //loop until all leters are masked
{
mask[i] = i + 127;
printf("\nYour compressed word/sentence is %c ", mask[i]);
}
return;
}
void extract(void) {
printf("\n-------------------------\n");
printf(" EXTRACT ");
printf("\n-------------------------\n");
return;
}
You are scanning only one single character, and as "%c" doesn't skip white-space this is the newline character terminating the previous input.
You instead want to read in a string, and to be on the safe side you should add the maximum length to read to: "%199s" (note: one less than array size to leave space for the terminating null character):
scanf_s("%199s", txt);
Note, too, that as txt is an array it decays to a pointer automatically when being passed to a function (see above); taking the address of (&txt) produces a pointer with the same value, but of a different type: char(*)[200]. This pointer is not compatible to neither %c nor %s format specifier, thus you actually produce *undefined behaviour!
Note, too, that scanf_s (any function from scanf family) will stop reading at the first whitespace – a sentence might, though contain multiple words separated by whitespace. You'd just read the first one of them, though. So you might want to drop scanf for this input entirely in favour of e.g. fgets:
fgets(txt, sizeof(txt), stdin);
Note, here, too, that the previous scanf("%d", ...) did not consume the terminating newline, so you'll need to ignore that, e.g. by a preceding call to getchar.
Crediting this last point to Jonathan Leffler who hinted to in his comment to the question)
I am trying to take in user input with spaces and store it in an array of characters.
After, I want to take in a single character value and store it as a char.
However, when I run my code, the prompt for the character gets ignored and a space is populated instead. How can I take in an array of chars and still be allowed to prompt for a single character after?
void main()
{
char userIn[30];
char findChar;
printf("Please enter a string: ");
scanf("%[^\n]s", userIn);
printf("Please enter a character to search for: ");
scanf("%c", &findChar);
//this was put here to see why my single char wasnt working in a function I had
printf("%c", findChar);
}
scanf("%c", &findChar); reads the next character pending in the input stream. This character will be the newline entered by the user that stopped the previous conversion, so findChar will be set to the value '\n', without waiting for any user input and printf will output this newline without any other visible effect.
Modify the call as scanf(" %c", &findChar) to ignore pending white space and get the next character from the user, or more reliably write a loop to read the read and ignore of the input line.
Note also that scanf("%[^\n]s", userIn); is incorrect:
scanf() may store bytes beyond the end of userIn if the user types more than 29 bytes of input.
the s after the ] is a bug, the conversion format for character classes is not a variation of the %s conversion.
Other problems:
void is not a proper type for the return value of the main() function.
the <stdio.h> header is required for this code.
Here is a modified version:
#include <stdio.h>
int main() {
char userIn[30];
int c;
char findChar;
int i, found;
printf("Please enter a string: ");
if (scanf("%29[^\n]", userIn) != 1) {
fprintf(stderr, "Input failure\n");
return 1;
}
/* read and ignore the rest of input line */
while ((c = getchar()) != EOF && c != '\n')
continue;
printf("Please enter a character to search for: ");
if (scanf("%c", &findChar) != 1) {
fprintf(stderr, "Input failure\n");
return 1;
}
printf("Searching for '%c'\n", findChar);
found = 0;
for (i = 0; userIn[i] != '\0'; i++) {
if (userIn[i] == findChar) {
found++;
printf("found '%c' at offset %d\n", c, i);
}
}
if (!found) {
printf("character '%c' not found\n", c);
}
return 0;
}
scanf("%[^\n]s", userIn); is a bit weird. The s is guaranteed not to match, since that character will always be \n. Also, you should use a width modifier to avoid a buffer overflow. Use scanf("%29[^\n]", userIn); That alone will not solve the problem, since the next scanf is going to consume the newline. There are a few options. You could consume the newline in the first scanf with:
scanf("%29[^\n]%*c", userIn);
or discard all whitespace in the next call with
scanf(" %c", &findChar);
The behavior will differ on lines of input that exceed 29 characters in length or when the user attempts to assign whitespace to findChar, so which solution you use will depend on how you want to handle those situations.
While trying different things with getchar I figured out that it usually only safes on character in an variable. Somehow when I use a while loop the behaviour changes and it returns more characters if the input is more than one.
Here is my example code for "normal" behaviour. putchar() return only one character even more are put in:
#include <stdio.h>
main() {
char c;
printf("Type a character in here: ");
c = getchar();
printf("You just typed : ");
putchar(c);
}
Somehow when I want to use it in a while loop the putchar() function returns more than one character if more are put in:
Here is the second part:
#include <stdio.h>
void print_input();
main() {
char c;
printf("Type a character in here: ");
c = getchar();
printf("You just typed : ");
putchar(c);
print_input();
}
void print_input() {
char ch = 'x';
while (ch != '#') {
ch = getchar();
putchar(ch);
}
return;
}
Additional question:
While running this in debugger the behaviour somehow is different than if I try this in runtime. Why is that so?
When you enter multiple chars and hit enter, the program will see that whole input (because it's line buffered). So multiple calls to getchar will return subsequent characters and remove them from the stream:
Try to play with this:
char c;
char d;
printf("Type chars in here: ");
c = getchar();
d = getchar();
printf("C: %c \n", c);
printf("D: %c", d);
This is duplicated with:
How is the "getchar()" function able to take multiple characters as input?
You can read more there.
Here first gets() is not working. if I add one more gets() function then from the two last one goes to work. how can I fix it?
CODE
#include<stdio.h>
#include<string.h>
int main(void)
{
short int choice;
char number[15];
do{
printf("\n\nAnswer: ");
scanf("%hd",&choice);
printf("\n");
if(choice==1)
{
printf("Enter the decimal number: ");
gets(number);
}
else
{
printf("Wrong input!.");
system("pause");
system("cls");
}
}while(choice!=1);
return 0;
}
Because the when the user pressed the enter key to give you the input for the scanf call, the enter key added a newline in the input buffer. And the gets call read that newline as an empty line.
One way to solve it is to use fgets to read the first input too, and use sscanf to parse it to a number:
...
printf("\n\nAnswer: ");
char input[64];
fgets(input, sizeof(input), stdin);
sscanf(input, "%hd", &choice);
printf("\n");
...
This make sure that the newline after the input is read and skipped.
Another way is to read one character at a time in a loop after the scanf call, until you have read the newline:
scanf("%hd", &choice);
int ch;
while ((ch = fgetc(stdin)) != EOF && ch != '\n')
{
// Empty
}
And a third way is to simply ask the scanf call to read and ignore white-space after the input:
scanf("%hd ", &choice);
// ^
// |
// Note space here
All of these methods have both pros and cons. You can try them all and use the one that works for you.
You need to skip the whitespace (i.e. the newline) following the number in the input buffer. This can be done by modifying the scanf to:
scanf("%hd ",&choice);
And use fgets(), since gets() is prone to buffer overflows.
New bee in C. This is my code (It replaces a character from a string):
#include <stdio.h>
#include <string.h>
#include <conio.h>
void main()
{
char str[100], r, ra;
printf("enter string");
gets(str);
int length;
length= strlen(str);
printf("length of string is %d",length);
printf("\nenter the the character that will replace");
scanf("%c",&r);
printf("where to replace\n b...begning\ne....ending\np....position");
scanf("%c",&ra);
int pos;
switch(ra)
{
case 'b' : str[1]=r; break;
case 'e' : str[length-1] = r; break;
case 'p' : printf("enter position");
scanf("%d",pos);
if(pos<1 || pos>length-1)
printf("please enter a position between 1 and %d",length-1);
else
str[pos]= r;
break;
}
printf("\n after replacing string is %s", str);
getche();
}
The problem is that the IDE is not compiling this part of the program, I know that I am doing some thing wrong, but can't figure out what? Need help please.
scanf("%c",&ra);
int pos;
switch(ra)
{
case 'b' : str[1]=r; break;
case 'e' : str[length-1] = r; break;
case 'p' : printf("enter position");
scanf("%d",pos);
if(pos<1 || pos>length-1)
printf("please enter a position between 1 and %d",length-1);
else
str[pos]= r;
break;
}
use scanf(" %c",&ra) insted of "%c". Because reading with "%c" give you a garbage value in ra.And that value is new line.
When you enter value in a you press something like p and then Enter key. This Enter key still remains in stdin stream.
Next time when you read in ra then the Enter key in stdin stream is returned in ra.
So for removing that Enter key you need to read like " %c".
scanf(" %c", &ra); // space before %c
Unlike most conversions, %c does not skip whitespace before converting a character. After the user enters the number, a carriage return/new-line is left in the input buffer waiting to be read -- so that's what the %c reads.. SO POST
And for the same reason your switch case is not working, since ra does not have the expected value
the problem is that the ide is not compiling this part of the program
Well, that's a strong accusation. Rather than assume that the compiler does decide not to compile part of the code (on a whim), it's a safer bet that your program's execution flow just does not enter that part as you expected.
In particular, scanf does not behave as you think it does. It reads from stdin, which is a buffered input stream. "Buffered" means that it does not provide your program with input until a newline in read, i.e. until the user presses return. But the scanf family of functions doesn't look for new lines, it treats the new-line character as a normal character. In your case, scanning "%c" tries to read any character from the input. The subsequent "%c" then reads the new line, so &ra really is '\n' in your switch statement.
I usually find working with direct input from the user difficult in C, but if you must prompt the user interactively, I suggest that you read in a whole line of input first with fgets and then analyse that line with sscanf. That gets rid of the seemingly out-of-sync input and also allows you to scan a line several times, perhaps for alternative input syntaxes.
So, here's a version of your code that uses this technique:
#include <stdio.h>
#include <string.h>
int main()
{
char str[100], r, ra;
char line[20];
int length;
int pos;
printf("enter string");
fgets(str, 100, stdin); // note: str includes trailing newline
length = strlen(str);
printf("length of string is %d\n", length);
printf("enter the the character that will replace:\n");
fgets(line, 20, stdin);
sscanf(line, " %c ",&r);
printf("where to replace\n");
printf("b...begning\ne....ending\np....position\n");
fgets(line, 20, stdin);
sscanf(line, " %c ", &ra);
switch (ra)
{
case 'b': str[1] = r;
break;
case 'e': str[length - 1] = r;
break;
case 'p': printf("enter position");
fgets(line, 20, stdin);
sscanf(line, "%d ", &pos);
if(pos < 1 || pos > length-1)
printf("please enter a position between 1 and %d",
length-1);
else
str[pos]= r; break;
}
printf("after replacing string is %s", str);
return 0;
}
There are still problems with your code, mainly to do with zero-based array indexing in C. I leave it to you to sort those out. Also, prefer the safer fgets(buf, len, stdin) over gets(str), which does not prevent buffer overflow. And your query for a position should take a pointer to the address of pos, not just pos. And please make a habit of putting the new-line character last in your printf strings. It makes for cleaner reading and matches the way that the buffered output stream works.
The program doesn't compile, the most likely reason is that you are using a compiler that supports C89 only (I guess it's Visual Studio), or you are using C89 mode.
In this code:
scanf("%c",&ra);
int pos;
switch(ra)
{
the variable pos is defined in the middle of a block, which is supported only since C99. The solution is to move all definitions up to the beginning of a block:
int main()
{
char str[100], r, ra;
int pos;
printf("enter string");
Use fgets() to replace gets(), use int main to replace void main. And fix the problem with using scanf that is covered by the other answers.