I have the following C code:
#include <stdio.h>
int main()
{
char* ptr;
printf("Enter the word: ");
gets(ptr);
printf("The input string is: ");
puts(ptr);
return 0;
}
It compiles and asks for the input, but after I enter the input, it takes a pause and exits. No further commands are executed or displayed. I am unable to understand the problem. Please help.
#include <stdio.h>
int main()
{
char str[100] = "";
printf("Enter the word: ");
fgets(str, sizeof str, stdin);
printf("The input string is: ");
printf("%s", str);
return 0;
}
Try to use fgets(). gets() is dangerous to use because gets() is inherently unsafe, because it copies all input from STDIN to the buffer without checking size. This allows the user to provide a string that is larger than the buffer size, resulting in an overflow condition.
puts is simpler than printf but be aware that the former automatically appends a newline. If that's not what you want, you can fputsyour string to stdout or use printf.
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 3 years ago.
In C:
I'm trying to get char from the user with scanf and when I run it the program don't wait for the user to type anything...
This is the code:
char ch;
printf("Enter one char");
scanf("%c", &ch);
printf("%c\n",ch);
Why is not working?
The %c conversion specifier won't automatically skip any leading whitespace, so if there's a stray newline in the input stream (from a previous entry, for example) the scanf call will consume it immediately.
One way around the problem is to put a blank space before the conversion specifier in the format string:
scanf(" %c", &c);
The blank in the format string tells scanf to skip leading whitespace, and the first non-whitespace character will be read with the %c conversion specifier.
First of all, avoid scanf(). Using it is not worth the pain.
See: Why does everyone say not to use scanf? What should I use instead?
Using a whitespace character in scanf() would ignore any number of whitespace characters left in the input stream, what if you need to read more inputs? Consider:
#include <stdio.h>
int main(void)
{
char ch1, ch2;
scanf("%c", &ch1); /* Leaves the newline in the input */
scanf(" %c", &ch2); /* The leading whitespace ensures it's the
previous newline is ignored */
printf("ch1: %c, ch2: %c\n", ch1, ch2);
/* All good so far */
char ch3;
scanf("%c", &ch3); /* Doesn't read input due to the same problem */
printf("ch3: %c\n", ch3);
return 0;
}
While the 3rd scanf() can be fixed in the same way using a leading whitespace, it's not always going to that simple as above.
Another major problem is, scanf() will not discard any input in the input stream if it doesn't match the format. For example, if you input abc for an int such as: scanf("%d", &int_var); then abc will have to read and discarded. Consider:
#include <stdio.h>
int main(void)
{
int i;
while(1) {
if (scanf("%d", &i) != 1) { /* Input "abc" */
printf("Invalid input. Try again\n");
} else {
break;
}
}
printf("Int read: %d\n", i);
return 0;
}
Another common problem is mixing scanf() and fgets(). Consider:
#include <stdio.h>
int main(void)
{
int age;
char name[256];
printf("Input your age:");
scanf("%d", &age); /* Input 10 */
printf("Input your full name [firstname lastname]");
fgets(name, sizeof name, stdin); /* Doesn't read! */
return 0;
}
The call to fgets() doesn't wait for input because the newline left by the previous scanf() call is read and fgets() terminates input reading when it encounters a newline.
There are many other similar problems associated with scanf(). That's why it's generally recommended to avoid it.
So, what's the alternative? Use fgets() function instead in the following fashion to read a single character:
#include <stdio.h>
int main(void)
{
char line[256];
char ch;
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
ch = line[0];
printf("Character read: %c\n", ch);
return 0;
}
One detail to be aware of when using fgets() will read in the newline character if there's enough room in the inut buffer. If it's not desirable then you can remove it:
char line[256];
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
line[strcpsn(line, "\n")] = 0; /* removes the trailing newline, if present */
This works for me try it out
int main(){
char c;
scanf(" %c",&c);
printf("%c",c);
return 0;
}
Here is a similiar thing that I would like to share,
while you're working on Visual Studio you could get an error like:
'scanf': function or variable may be unsafe. Consider using scanf_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS
To prevent this, you should write it in the following format
A single character may be read as follows:
char c;
scanf_s("%c", &c, 1);
When multiple characters for non-null terminated strings are read, integers are used as the width specification and the buffer size.
char c[4];
scanf_s("%4c", &c, _countof(c));
neither fgets nor getchar works to solve the problem.
the only workaround is keeping a space before %c while using scanf
scanf(" %c",ch); // will only work
In the follwing fgets also not work..
char line[256];
char ch;
int i;
printf("Enter a num : ");
scanf("%d",&i);
printf("Enter a char : ");
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
ch = line[0];
printf("Character read: %c\n", ch);
try using getchar(); instead
syntax:
void main() {
char ch;
ch = getchar();
}
Before the scanf put fflush(stdin); to clear buffer.
The only code that worked for me is:
scanf(" %c",&c);
I was having the same problem, and only with single characters. After an hour of random testing I can not report an issue yet. One would think that C would have by now a bullet-proof function to retrieve single characters from the keyboard, and not an array of possible hackarounds... Just saying...
Use string instead of char
like
char c[10];
scanf ("%s", c);
I belive it works nice.
Provides a space before %c conversion specifier so that compiler will ignore white spaces. The program may be written as below:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char ch;
printf("Enter one char");
scanf(" %c", &ch); /*Space is given before %c*/
printf("%c\n",ch);
return 0;
}
You have to use a valid variable. ch is not a valid variable for this program. Use char Aaa;
char aaa;
scanf("%c",&Aaa);
Tested and it works.
I have the following program that takes as input the batsman names and their scores and prints the batsman with the highest score. I have written the following algorithm and it works. But the only problem I am facing is that, the newline character is getting displayed on the screen after the input has been gotten from the user.
#include<stdio.h>
#include<stdlib.h>
#include<limits.h>
#include<string.h>
int main()
{
int n;
char bat[100],maxs[100];
int score,max=INT_MIN;
scanf("%d",&n);
while(n--)
{
scanf("%99[^,],%d",bat,&score);
if(score>max)
{
max=score;
strcpy(maxs, bat);
}
}
printf("%s",maxs);
}
I have no clue of where the newline is coming from? Please see the output shown below. Any help is appreciated.
Imagine the following program:
#include <stdio.h>
int main() {
int a;
scanf("%d", &a);
char string[100];
scanf("%99[^,]", string);
printf("-----\n");
printf("%s", string);
}
Now execution could look like:
10 # %d scans 10 and leaves the newline in input
string, # then %99[^,] reads from the newline including it up until a ,
-----
string
How can I resolve this so that the newline is removed?
Read the newline. A space character in scanf ignores all whitespace charactesrs.
scanf(" %99[^,]", string);
You could ignore a single newline character, if you want to be "exact":
scanf("%*1[\n]%99[^,]", string);
You're getting a newline there because scanf() requires you to hit enter to proceed. This enter then gets stored in the string as well. You could remove newlines at the end or beginning with something like this (source here):
void remove_newline_ch(char *line)
{
int new_line = strlen(line) -1;
if (line[new_line] == '\n')
line[new_line] = '\0';
}
#include <stdio.h>
int main(){
char phrase [100]={0};
printf("Write a sentence: \n");
scanf ("%s", phrase);
printf("%s\n", phrase);
}
The purpose of this program is to print a sentence from the user. The problem is that there is something wrong with the ' ' character and i dont know how to solve it.
During the execution, the sentence is only printed until the first empty space, for some reason. but why?
As pointed out in comments, the %s conversion specifier in scanf can't read strings with white spaces; it uses white spaces as delimiters. You can use fgets :
#include <stdio.h>
#include <string.h>
int main() {
char aLine[100];
fgets(aLine, 100, stdin);
aLine[strlen(aLine)-1] = '\0'; // trimming the last \n
printf("%s\n", aLine);
return 0;
}
I tried using scanf twice for scanning a string and then scanning a char. It scans string first and does not execute the second scanf. When I use both %s and %c in a single scanf it works perfectly. Can you tell me why this happens?
#include<stdio.h>
int main()
{
char s[100],ch;
scanf("%s",s);
scanf("%c",&ch); //this does not work
printf("%s %c",s,ch);
return 0;
}
another program which works
#include<stdio.h>
int main()
{
char s[100],ch;
scanf("%s %c",s,&ch); //this works!
printf("%s %c",s,ch);
return 0;
}
Please add a space before %c in scanf().
There is a newline character after the string is read so this is being taken by %c
#include<stdio.h>
int main()
{
char s[100],ch;
scanf("%s",s);
scanf(" %c",&ch);
printf("%s %c",s,ch);
return 0;
}