the idea is that i type in a sentence and store it as a string... then i can choose a letter and change it to a different letter
int main(int argc, char *argv[])
{
char string[100];
char newLetter;
char oldLetter;
int i = 0;
printf("Please enter your sentence : ");
gets(string);
printf("\n\nWord is : %s" , string );
printf("\n\nTarget : ");
scanf("%s", &oldLetter);
printf("Replace with : ");
scanf("%s", &newLetter);
for ( i = 0; i < sizeof(string); i++)
{
if (string[i] == oldLetter)
{
string[i] = newLetter;
break;
}
}
printf("\n\nWord is : %s" , string );
system("PAUSE");
return 0;
}
any help in where I've gone wrong is appreciated
eg.
input could be - yellow lorry red lorry
then target - r
change to - t
output - yellow lotty ted lotty
Change the %s in the scanning of the two letters to %c and the code will run flawlessly.
First things first, since you are aiming to store a character to char oldLetter and char newLetter via scanf, you should be using the format specifier %c instead of the %s.
However, that won't be enough, because of the following: When you use functions like scanf or gets, you prompt user to input characters to the stdin stream. stdin stream is a buffered stream. You may think of it as a:
Stream of river that you;
drop characters into
each character you drop remains inside the river
until something takes them out
When scanf comes, and you type, say, A and then press the enter key, you put the following characters in the stream:
'A' '\n'
Where \n is the new-line character. With the enter key-press, you also inform the scanf that you're done. scanf then starts issuing the stdin buffer, let's see... 'A', a proper value for %c character. It takes that one out, leaves \n behind.
Then the next scanf comes, seeks for a %c in stream, finds the \n ready, takes that out. This is what you wouldn't want. Two ways to prevent it:
use fflush( stdin ); after the scanf calls, or
use while( getchar( ) != '\n' ); after the scanf calls
to dismiss/discard the remaining characters in the buffer.
And get rid of that break; if you want to replace each occurrence and not just the first one.
You can do it right, or you can do it wrong and hope that the person doesn't enter the right series of keystrokes or pipe in a file that doesn't end in newline, to break it ... up to you
Related
I know that adding a space in front of %c in scanf() will scan my second character; however, if two letters were inputted in the first character, it will input the second letter into the second character. How do I scan a single character only?
#include <stdio.h>
int main(void)
{
char firstch, secondch;
printf("Enter your first character: ");
scanf("%c", &firstch);
printf("Enter your second character: ");
scanf(" %c", &secondch);
printf("\n Fisrt character : %c \n Second character : %c \n", firstch, secondch);
return 0;
}
This is my result after running:
Enter your first character: ab
Enter your second character:
First character : a
Second character : b
I only want to read the first character 'a', but the second letter 'b' was inputted right away before I enter my second character.
When you are reading a line of user-input, use a line-oriented input function like fgets() or POSIX getline(). That way the entire line of input is read at once and you can simply take the first character from the line. Say you read a line into the array used as buffer called buf, e.g.
#define MAXC 1024 /* if you need a constant, #define one (or more) */
int main (void) {
char buf[MAXC]; /* buffer to read each line into */
You can simply access the first character as buf[0], or since buf[0] is equivalent to *(but + 0) in pointer notation, you can simply use *buf to get the first character.
As a benefit, since all line-oriented functions read and include the '\n' generated by the user pressing Enter after the input, you can simply check if the first character is '\n' as a way of indicating end-of-input. The user simply presses Enter alone as input to indicate they are done.
Using a line-oriented approach is the recommended way to take user input because it consumes and entire line of input each time and what remains in stdin unread doesn't depend on the scanf conversion specifier or whether a matching failure occurs.
Using " %c%*[^\n]" is not a fix-all. It leaves the '\n' in stdin unread. That's why you need the space before " %c". Where it is insidious is if your next input uses a line-oriented function after your code reading characters is done. Unless you manually empty the '\n' from stdin, before your next attempted line-oriented input, that input will fail because it will see the '\n' as the first character remaining in stdin.
A short example using fgets() for a line-oriented approach would be:
#include <stdio.h>
#define MAXC 1024 /* if you need a constant, #define one (or more) */
int main (void) {
char buf[MAXC]; /* buffer to read each line into */
for (;;) { /* loop continually */
fputs ("enter char: ", stdout); /* prompt for input */
/* read/validate line, break on EOF or [Enter] alone */
if (!fgets (buf, sizeof buf, stdin) || *buf == '\n')
break;
printf (" got: %c\n\n", *buf); /* output character read */
}
}
Where you simply take input continually isolating the first character as the value you want until the user presses Enter alone to break the read-loop.
Example Use/Output
$ ./bin/fgetschar
enter char: a
got: a
enter char: ab
got: a
enter char: a whole lot of stuff you don't have to deal with using fgets()
got: a
enter char: banannas
got: b
enter char: cantelopes
got: c
enter char:
Look things over and let me know if you have further questions.
Using a space before the %c will skip whitespace before scanning the next non-whitespace character. %c itself just scans a single character -- the next character in the input after whatever else was scanned or skipped previously.
So the question is, what do you want to do? Do you want to skip over all extraneous input on the line after the first character (up to newline?) fgets or scanf("%*[^\n]"); scanf("%c"); will do that (but be careful -- if firstch was itself a newline, this will skip the next line.) Do you want to check the input and make sure it is exactly one character on a line? If so, use fgets (not scanf) and check that the line read is exactly two characters (a character and a newline). Or perhaps you really want to read keystrokes without having the user hit Enter after esch one? That requires changing the input source setup, which is OS dependent.
I'm still new to C coding, and I've found a suitable answer to my problem by using scanf("%*[^\n]");
#include <stdio.h>
int main(void)
{
char firstch, secondch;
printf("Enter your first character: ");
scanf(" %c%*[^\n]", &firstch);
printf("Enter your second character: ");
scanf(" %c%*[^\n]", &secondch);
printf("\n First character : %c \n Second character : %c \n", firstch,
secondch);
return 0;
}
Results after running:
Enter your first character: ab
Enter your second character: c
First character : a
Second character : c
Thanks to #Eraklon #Chris Dodd #David C. Rankin
I wrote a program to replace a letter in a string. Although it has no error, the output is not as expected. Please help me with it.
#define _CRT_SECURE_NO_DEPRECATE
#include<stdio.h>
#include<string.h>
void replace(char s,char d);
char a[100];
int main()
{
char b,r;
printf("enter the string\n:");
gets(a);
printf("enter the the letter to be replaced\n:");
scanf("%c", &b);
printf("enter the letter to be replaced with\n:");
scanf("%c", &r);
replace(b,r);
}
void replace(char s, char d)
{
int i,f=0;
for (i = 0; a[i] != '\0'; i++)
{
if (a[i] == s)
{
a[i] = d;
f = 1;
}
}
if (f == 0)
{
printf("letter not found");
}
}
Output
enter the string
:hello every one
enter the the letter to be replaced
:e
enter the letter to be replaced with
:letter not found
I wanted to replace e with o but I am not able to give the input for word to be replaced
UPDATE
Use this loop to get rid of the input buffer problem when using scanf
but I am not sure how to implement it on my program need help
void
clear(void)
{
while ( getchar() != '\n' )
;
}
The scanf() function skips over initial whitespace characters when you read in strings using the %s specifier, but it does not do this when your read chars with the %c specifier. The gets() function that you use (which you should never ever ever use ever) reads through the newline, and discards it. So your first call to scanf() has a clean input stream. When you call scanf() the first time, a value is read into the variable b, but the trailing newline is left behind in the input stream. Then, when you try to read the next value, scanf() picks up this newline, instead of the value that you want to enter.
One fix for this is to discard any unwanted characters from the input stream like this:
while (getchar() != '\n')
continue; // discard unwanted characters
You can also test for the EOF character in the conditional expression if you really want to be careful. One virtue of this approach is that, no matter how many characters the user enters at your second prompt, only the first is taken, and the remaining characters through the newline are discarded. Since there is nothing left in the input stream, scanf() has to wait for the user to enter something at your third prompt. You should place this code after each call to scanf() to make sure that the input stream is clear.
Now, gets() is a terrible and unsafe function begging for buffer overflows, because it doesn't check to see if there is enough memory allocated for the string it is getting. Instead, use fgets(). This function takes an argument that specifies the maximum number of characters to read, including the null-terminator. fgets() also reads the newline character into the string, so you have to dispose of that yourself if you don't want it. Here are the modifications you need to make:
int i = 0;
...
char b,r;
printf("enter the string\n:");
fgets(a, 100, stdin);
while(a[i] != '\n' && a[i] != '\0') // remove newline
++i;
a[i] = '\0';
printf("enter the the letter to be replaced\n:");
scanf("%c", &b);
while (getchar() != '\n')
continue; // discard unwanted characters
printf("enter the letter to be replaced with\n:");
scanf("%c", &r);
while (getchar() != '\n')
continue; // discard unwanted characters
replace(b,r);
printf("%s\n", a);
...
I added a final printf() to display the changed string.
I am using GCC to compile my C code.
My second scanf is not stopping to get the input.
It only reads in the first scanf and prints the two statements, one with what I entered in string and the other is just blank.
int main (void) {
setvbuf(stdout, NULL, _IONBF, 0);
char string[25] = {'\0'};
char c;
scanf(" %s", string);
scanf(" o%c", &out);
printf("Input is : %s \n\n", string);
printf("Out is: %c", out);
return 0;
}
Instead of getting
Input is whatever I typed and a prompt to enter a char for out
I got output as shown below
Input is : whatever i typed
Out is:
The program terminates. Can someone help. I've done some research and tried to put a space before %c for out and for string and still nothing happened.
You haven't defined out. And c is unused here. Having said that
Change
scanf(" o%c", &out); //What is that o in here? Is it a typo?
to
scanf(" %c", &out);
If your terminal uses line-buffered input
scanf(" %s", string);
can read the input till the first white-space. So the input buffer from the white-space is unused which is available for the next scanf which automatically starts reading from the buffer. So if you enter a string with spaces, the white-space will be assigned to the character out in your case
Change first scanf like below to clear the buffer:
if( scanf(" %s", string) == 1)
{
while(getchar()!='\n')
continue;
}
Also you might wish to replace
scanf(" %s", string);
with
fgets(string,25,stdin);
/* Use 26 if you actually wish to have max 25 characters
* ie char string[26]={'\0'}
*/
fgets has the advantage that it can read the white spaces and the newline charcter '\n' and it will automatically trim the output in case of an overflow.
I'm newcomer to C and I am stuck. I want to write simple program, which will take input from keyboard and output it if it isn't an 'exit' word. I've tried few different approaches and none of them works. Almost in all cases I get infinite output of the first input.
Here is one of my approaches:
#include <stdio.h>
int main() {
char word[80];
while (1) {
puts("Enter a string: ");
scanf("%79[^\n]", word);
if (word == "exit")
break;
printf("You have typed %s", word);
}
return 0;
}
I thought after it finish every loop it should give me prompt again, but it doesn't.
What I am doing wrong.
Please if you know give me some advice.
Thanks in advance. Really, guys I will be so happy if you help me to understand what I am doing wrong.
Oh, by the way I've noticed that when I typed some word and press 'Enter', the result string also include Enter at the end. How can I get rid of this ?
Improper string compare - use strcmp().
if (word == "exit") simply compares 2 address: the address of the first char in word and the address of the first char in string literal "exit". Code needs to compare the content beginning at those addresses: strcmp() does that.
Left-over '\n' from the previous line's Enter. Add a space to scanf() format to consume optional leading white-space. Also check scanf() results.
scanf() specifiers like "%d", "%u" and "%f" by themselves consume optional leading white-space. 3 exceptions: "%c", "%n" and "%[".
Add '\n' at end of printf() format. # Matt McNabb
#include <stdio.h>
int main() {
char word[80];
while (1) {
puts("Enter a string: ");
// v space added here
if (scanf(" %79[^\n]", word) != 1)
break; // Nothing saved into word or EOF or I/O Error
if (strcmp(word, "exit") == 0)
break;
printf("You have typed %s\n", word);
}
return 0;
}
Nice that OP used a proper width limited value of 79 in scanf()
Oh, by the way I've noticed that when I typed some word and press 'Enter', the result string also include Enter at the end. How can I get rid of this ?
This is because you don't output a newline after printf("You have typed %s", word);. The next statement executed is puts("Enter a string: "); . So you will see You have typed helloEnter a string:. To fix this, change to printf("You have typed %s\n", word);
As others have mentioned, use strcmp to compare strings in C.
Finally, the scanf format string "%79[^\n]" does not match a newline. So the input stream still contains a newline. Next time you reach this statement the newline is still in the stream , and it still doesn't match because you specifically excluded newlines.
You will need to discard that newline (and any other input on the line) before getting the next line. One way to do that is to change the input to scanf("%79[^\n]%*[^\n]", word); getchar(); That means:
Read up to 79 non-newlines
Read all the non-newline things , and don't store them
Read a character (which must be a newline now) and don't store it
Finally it would be a good idea to check the return value of scanf so that if there is an error then you can exit your program instead of going into an infinite loop.
The specifier [^\n] will abort scanf if the next character is a newline (\n), without reading the newline. Because of that, the scanf calls after the first one won't read any input.
If you want to read single words, use the %79s specifier and the following code to remove the \n at the end of your string:
if(word[strlen(word)]=='\n')
word[strlen(word)]='\0';
If you want to read whole lines, you can remove the newline from the input buffer this way:
char line[80];
int i;
while(1)
{
puts("Enter a string:");
i=-1;
scanf("%79[^\n]%n",line,&i);
//%n returns the number of characters read so far by the scanf call
//if scanf encounters a newline, it will abort and won't modify i
if(i==-1)
getchar(); //removes the newline from the input buffer
if(strcmp(line,"exit")==0)
break;
printf("You have typed %s\n",line);
}
return 0;
It is better to clear (to have a reproducible content) with memset(3) the memory buffer before reading it, and you should use strcmp(3) to compare strings. Also, consider using fflush(3) before input (even if it is not actually necessary in your case), don't forget to test result of scanf(3), also most printf(3) format control strings should end with a \n -for end-of-line with flushing- so:
#include <stdio.h>
int main() {
char word[80];
while(1) {
puts("Enter a string: ");
memset (word, 0, sizeof(word)); // not strictly necessary
fflush(stdout); // not strictly necessary
if (scanf("%79[^\n]", word)<=0) exit(EXIT_FAILURE);
if (!strcmp(word,"exit"))
break;
printf("You have typed %s\n", word);
};
return 0;
}
I would suggest reading a whole line with fgets(3) and getting rid of its ending newline (using strchr(3)). Also read about getline(3)
Don't forget to compile with all warnings and debug info (e.g. gcc -Wall -g) and learn how to use the debugger (e.g. gdb)
Your first problem is that you can't compare a string with '=='. So:
if (word == "exit")
should be
if ( strncmp( word, "exit", 4 ) == 0 )
(You could also use strncmp( word, "exit", strlen(word) ) if you know that word is zero-terminated and safe from bad values. There's a few other options also.)
Your second problem is that scanf() is not consuming the input, probably because it's not matching what you've told it to expect. Here is a good explanation of how to do what you want to do:
http://home.datacomm.ch/t_wolf/tw/c/getting_input.html
i use a scanf to get user input but if i press enter, the cursor will flash to next line~
what function should i use instead of scanf if i want the program will terminated if the users only press enter without keying any thing?
thanks
Scanf reads until the next token -- it doesn't really care about newlines at all (just considers them to be whitespace, like spaces or tabs).
Instead, use a line-reading function like fgets.
you can use gets() function in this way :
#include <stdio.h>
#define MAX_INPUT_CHAR 100
int main( ) {
char str[MAX_INPUT_CHAR ];
printf( "Enter a value :");
gets( str );
return 0;
}
Function char *gets(char *s) reads a line from stdin into the buffer pointed to by
s until either a terminating newline or EOF (End of File)
Pay attention gets() is not safe. You can't now apriori how much character it will read. This can cause some security problem and eventually crash.
Just use this code :
char ch;
printf("Enter Your Sentence \n");
while (1)
{
ch = getch();
if ((int)ch == 13)
break;
printf("%c", ch);
}
I disagree:
char * x;
scanf("%s",x);
printf("Inserted line: %s\n",x);
Will store in x only the characters till the return key.
I build it and runned it and it works.