I am trying to print the ASCII values of 3 char-type characters. When I input the first char it doesn't print the value of the char. After the first char it starts to give the value.
#include <stdio.h>
int main() {
char ch;
int t;
while(t < 3){
scanf("%c\n", &ch);
printf("%c - %d\n", ch,ch);
t++;
}
}
http://i54.tinypic.com/2mdqb7d.png
Variable t is not automatically initialized to 0 by compiler. So You need to initialize t with 0. If printf doesn't print immediately it means the data is buffered. If you want to see immediatley you may consider flushing stdout right after printf.
I saw this several times, and don't know the root cause, but solution that works is:
scanf("\n%c", &ch);
It probably has something to do with buffered end of line character.
Related
as a homework assignment for my computing 1 college course, my professor has given me the task of having the user input a string of characters into the terminal, taking that string, adding it into an array, then printing the array and printing the array backwards. I think that I know of a way to print the array backwards, however, I cannot come up with a way to read from the terminal and add the characters from the terminal to an array. I have tried doing the following:
char ch;
for (int i = 0; i <= 80 || str[i] == '\n'; ++i) {
scanf_s("%c", &str[i]);
}
I am wondering if someone could explain to me why this section of code does not operate as expected, and if someone could give me some other ideas to try. Thank you.
You are using scanf_s with %c specifier incorrectly.
Please take notice of compiler warnings, there is a size argument missing.
Microsoft's scanf_s is not a direct replacement for scanf.
Unlike scanf ... scanf_s ... requires the buffer size to be specified for all input parameters of type c, C, s, S, or string control sets that are enclosed in []. The buffer size in characters is passed as an additional parameter immediately following the pointer to the buffer or variable.
scanf_s("%c", &str[i], 1);
You might also want to filter out any newline which may have been left in the buffer, with
scanf_s(" %c", &str[i], 1);
notice the added space.
Why your code is showing this type of behaviour...
use scanf instead of scanf_s
the conditions you have provided in the for loop are wrong
#include <stdio.h>
int main()
{
char ch;
char str[1000];
int i;
for (i = 0; i <= 80 ; i++)
{
scanf("%c", &str[i]);
if(str[i]=='\n')
{
str[i]='\0';
break;
}
}
printf(str);
}
I could show you the same task in simple manner. I have tried to answer your question in your way. That's why it may seem complicated.
#include <stdio.h>
#define MAX 25
int main()
{
char buf[MAX];
fgets(buf, MAX, stdin);
printf("%s\n", buf);
return 0;
}
fgets- Reads until new line character encountered or maximum limit of character array.
I am writing a program to count the occurrence of '2' followed by '1' in a sting.
I dynamically allocated string
Code is:
#include <stdio.h>
#include<stdlib.h>
int penalty_shoot(char* s){
int count=0,i=0;
while(s[i]!='\0'){
if(s[i]=='2')
if(s[i+1]=='1')
count++;
i++;
}
return count;
}
int main() {
int t;
int i=0;
scanf("%d",&t); //t is for number of test cases.
while(t--){
char *str, c;
str = (char*)malloc(1*sizeof(char));
while(c = getc(stdin),c!='\n')
{
str[i] = c;
i++;
str=realloc(str,i*sizeof(char));
}
str[i] ='\0';
printf("%s\n",str);
printf("%d\n",penalty_shoot(str));
free(str);
str=NULL;
i=0;
}
return 0;
}
Input is :
3
101201212110
10101
2120
I am facing 2 problems:
1) I feel the dynamic allocation is not working fine.I wrote the code for dynamic allocation seeing various codes on stackoverflow . (Can anyone suggest some changes.)
2) The code isn't reading '2120' as the 3rd input.
(why is it so ?)
Three errors:
Not checking for EOF:
Change while(c = getc(stdin),c!='\n') to while(c=getc(stdin),c!='\n'&&c!=EOF)
Reallocating with the wrong number of bytes:
Change str=realloc(str,i*sizeof(char)); to str=realloc(str,(i+1)*sizeof(char));
After taking one character input we increment i (i++), so the next character will be stored at the ith position. Now, in order to store the character at ith position, the length of the character array must be i+1. So, we realloc with i+1.
Just for the sake of brevity, as suggested by Basile, you
might as well do this:
Change str=realloc(str,(i+1)*sizeof(char)); to str=realloc(str,i+1);
Why? Because sizeof char is 1 byte
Not consuming the \n after inputting t:
Change scanf("%d",&t); to scanf("%d ",&t); or scanf("%d\n",&t);
scanf("%d ",&t); or scanf("%d\n",&t);.
Either of them works. Why, you ask? Read this explanation taken from another SO answer here:
An \n - or any whitespace character - in the format string consumes
an entire (possibly empty) sequence of whitespace characters in the
input. So the scanf only returns when it encounters the next
non-whitespace character, or the end of the input stream.
Tested here.
you can use scanf("%d ", &t); when user input to test
then just before second while loop, which condition should be c != '\n' write c = getchar();
and then make sure you create a char variable, i called mine clear, that receives 0 so when you loop after initiating your string you write c = clear; and under it c = getchar() again. and when you use realloc make sure you make it bigger by (i+1) since char is only the size of 1 byte.
we create the clear variable in order to clear the buffer.
it worked for me. make sure you insert the string all at once.
Why is this while loop not exiting ?
I want to input a series of number from users and then use them one by one for further processing. I do not have the size of numbers list. Can someone please tell me how to do this ?
#include<stdlib.h>
#include<stdio.h>
int main()
{
int x;
scanf("%d",&x);
while(x!='\n')
{
printf("%d",x);
scanf("%d",&x);
}
return 0;
}
Example:
Input:
3 5 6
Output:
3 5 6
Actual Output:
3 5 6
but loop doesn't exit
The %d conversion specifier never stores the \n into the buffer. It reads up to the \n in the buffer and leaves it there. So, this is essentially an infinite loop.
You may need to use a format specifier like %c which actually reads and stores the \n.
Although, if I may suggest, try to make use of getchar(), which reads the next character from the standard input and returns the value as an int. However, for an input of digits, you need to parse and store the integer and/or float value accordingly, but it will certainly provide you more flexibility.
What about this?
while(scanf("%d", &x) != EOF) {
printf("%d",x);
if (getchar() == '\n') break;
}
Why not use argv and strtol ?
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv) {
int i = 0, x;
for(i = 1; i < argc; i++) {
x = strtol(argv[i],NULL,10);
printf("%d\n",x);
}
return 0;
}
while(scanf("%d",&x) != EOF) {
printf("%d",x);
}
...would do it. Don't ignore scanf returns. The reason your code is looping is you're really comparing x against decimal value of 10 ('\n'). Also, please remember that you asked scanf to give you integers and that's what it gives you, it will not give you newlines, even if it could.
Edited, so that I added comparison to EOF specifically.
You need to remember that scanf scans a stream, in this case the input stream, it will not return until that stream is done.
You should really use strings and sscanf() for what you want to do.
i am a learner of 'C' and written a code, but after i compile it, shows a Debug Error message, here is the code:
#include<stdio.h>
void main()
{
int n,i=1;
char c;
printf("Enter Charecter:\t");
scanf("%s",&c);
printf("Repeat Time\t");
scanf("%d",&n);
n=n;
while (i <= n)
{
printf("%c",c);
i++;
}
}
Pls tell me why this happens and how to solve it
The scanf("%s", &c) is writing to memory it should not as c is a single char but "%s" expects its argument to be an array. As scanf() appends a null character it will at the very least write two char to c (the char read from stdin plus the null terminator), which is one too many.
Use a char[] and restrict the number of char written by scanf():
char data[10];
scanf("%9s", data);
and use printf("%s", data); instead of %c or use "%c" as the format specifier in scanf().
Always check the return value of scanf(), which is the number of successful assignments, to ensure subsequent code is not processing stale or uninitialized variables:
if (1 == scanf("%d", &n))
{
/* 'n' assigned. 'n = n;' is unrequired. */
}
scanf("%s",&c); should be scanf("%c",&c);
The %s format specifier tells scanf you're passing a char array. You're passing a single char so need to use %c instead.
Your current code will behave unpredictably because scanf will try to write an arbitrarily long word followed by a nul terminator to the address you provided. This address has memory allocated (on the stack) for a single char so you end up over-writing memory that may be used by other parts of your program (say for other local variables).
I'm not sure you understood the answer to your other question: Odd loop does not work using %c
These format specifiers are each used for a specific job.
If you want to get a:
character from stdin use %c.
string (a bunch of characters) use %s.
integer use %d.
This code:
char c;
printf("Enter Character:\t");
scanf("%c",&c);
Will read 1 character from stdin and will leave a newline ('\n') character there. So let's say the user entered the letter A in the stdin buffer you have:
A\n
The scanf() will pull 'A' and store it in your char c and will leave the newline character. Next it will ask for your int and the user might input 5. stdin now has:
\n5
The scanf() will take 5 and place it in int n. If you want to consume that '\n' there are a number of options, one would be:
char c;
printf("Enter Character:\t");
scanf("%c",&c); // This gets the 'A' and stores it in c
getchar(); // This gets the \n and trashes it
Here is a working version of your code. Please see inline comments in code for fixes:
#include<stdio.h>
void main()
{
int n,i=1;
char c;
printf("Enter Character:\t");
scanf("%c",&c);//Use %c instead of %s
printf("Repeat Time\t");
scanf("%d",&n);
n=n;//SUGGESTION:This line is not necessary. When you do scanf on 'n' you store the value in 'n'
while (i <= n)//COMMENT:Appears you want to print the same character n times?
{
printf("%c",c);
i++;
}
return;//Just a good practice
}
I am trying to take five character and 5 float input.
main()
{
char c[5];
float q[5];
int i;
for(i=0;i<5;i++)
{
printf("\n%d ",i);
scanf("%c",c+i);
scanf("%f",q+i);
}
}
But the output is absurd. After two sequential scans, it skips third scan and then again skips fifth scan.
I am not able to understand why is it showing such a behaviour.
I am working on gcc compiler.
Use of scanf is not recommended because of problems like this.
Instead use fgets to read the entire line and then use sscanf to extract what you want(a char or a float) from the line just read:
char line[MAX];
for(i=0;i<5;i++)
{
if( fgets(line,MAX,stdin) && sscanf(line,"%c", c+i)!=1 )
*(c+i) = 0;
if( fgets(line,MAX,stdin) && sscanf(line,"%f", q+i)!=1 )
*(q+i) = 0.0;
printf("%c %f\n",*(c+i),*(q+i));
}
To directly answer why the 3rd and every other scan "skips", it is the way scanf() and the %c format works. When there is a call to scanf(), you typically have to press enter to "submit" the input. Consequently that inserts a newline character into the stdin stream.
When the previous float scan got inputted, the newline is still left in the stream. When the character scan gets reached, that remaining newline character is read in since it fits effectively "skipping" the call.
You should use fgets() with sscanf() as codaddict suggests.
But as a quick fix, you could try adding a call to getchar() after the float scan to consume that newline character from the stream.
edit:
You're saying this doesn't work? (assuming you input the correct kinds of values, one per scanf call)
main()
{
char c[5];
float q[5];
int i;
for(i=0;i<5;i++)
{
printf("\n%d ",i);
scanf("%c",c+i);
scanf("%f",q+i);
getchar();
}
}
You should try this:
int main(){
char c[6];//char array size must be 6 to store five charecter
//as null-terminator('\0')will use the last one
float q[5];
int i;
for(i=0;i<5;i++){
printf("\n%d\n",i);fflush(stdout);
scanf("%c",&c[i]);fflush(stdin);//fflush(stdin) to clear input stream that
//next scanf() won't skip
scanf("%f",&q[i]);fflush(stdin);//fflush(stdin) to clear input stream that
//scanf() won't skip at new loop
}
return 0;
}
fflush() is not defined on an input stream, like stdin. Don't do it.
If you want to "read and discard until newline", then do:
int ch;
do {
ch = getchar();
} while (ch != EOF && ch != '\n');
Note that %c means "read the next character in the input stream, even if it's whitespace, then stop". %f means "read and discard whitespace, then try to read a float from the input stream, then stop."
Your code should be like this :
main()
{
char c[5];
float q[5];
int i;
for(i=0;i<5;i++)
{
printf("\n%d ",i);
scanf("%c",c+i);
while (getchar()!='\n');
scanf("%f",q+i);
while (getchar()!='\n');
}
}
the sentence while (getchar()!='\n'); search till the end of input, so it would not take '\n' as an input value for q+i.Also another while (getchar()!='\n'); after scanf q+i,since you use loop.
Problem in scanning a char value after a float value.
Solution is very simple!!!
instead of writting your code like this
scanf("%c",&...)
try this,
scanf(" %c",&...)
A Space before the"%c" will resolve the problem by neglecting the value of the Return(Enter) key when pressed after typing the value of the float as an input.
When a float value is scanned before a character value.
The value obtained by pressing the Return(Enter) key is collected in the following char variable. Using a space before(" %c",&...) discards the value collected of the Return(Enter) Key, Causing the Char value to be scanned in the next line. Thus solving The Scanning Float-Char problem.