How to scan initial spaces in C - c

I have a typical question it's not that how can I scan spaces using scanf but how to scan the initial spaces entered in a string
This is what I've done:
#include <stdio.h>
#include <string.h>
int main()
{
int n;
char a[10];
scanf("%d",&n);
scanf(" %[^\n]",a);
printf("%d",strlen(a));
return 0;
}
and when I run the program with following input:
aa bb//note there are two spaces before initial a
and the output is 6 but there are 8 characters i.e, 2 spaces followed by 2 a's followed by 2 spaces and then lastly 2 b's
I eve tried an own function.. but alas! the length is 6. Here's my function:
int len(char a[101])
{
int i;
for(i=0;a[i];i++);
return i;
}
What I think is that the initial 2 spaces are being ignored...or I might be wrong. It'd be great if someone could explain why the length of string is 6 and how can I make it 8 or accept all the 8 characters I mentioned above.
EDIT: this is my actual code
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int i,N,j,k;
char **ans,s[101];
scanf("%d",&N);
ans=(char **)calloc(N,sizeof(char*));
for(j=0,i=0;i<N;i++)
{
scanf(" %[^\n]",s);
printf("%d",strlen(s));
ans[i]=(char*)calloc(strlen(s),sizeof(char));
for(k=0,j=((strlen(s)/2)-1);j>=0;j--,k++)
{
ans[i][k]=s[j];
}
for(j=strlen(s)-1;j>=strlen(s)/2;k++,j--)
{
ans[i][k]=s[j];
}
}
for(i=0;i<N;i++)
{
printf("%s\n",ans[i]);
}
scanf("%d",&i);
return 0;
}

OP code should work as posted.
OP comments true code is using scanf(" %[^\n]",a); which fully explains the problem: the space in the format is consuming leading white-space.
To address other issues with scanf(), see following.
fgets() is the right tool.
Yet if OP insists on scanf()
how can I scan spaces using scanf but how to scan the initial spaces entered in a string?
char buf[100];
// Scan up to 99 non\n characters and form a string in `buf`
switch (scanf("%99[^\n]", buf)) {
case 0: buf[0] = '\0'; break; // line begins with `'\n`
// May want to check if strlen(buf)==99 to detect a long line
case 1: break; // Success.
case EOF: buf[0] = '\0'; break; // stdin is closed.
}
fgetc(stdin); // throw away the \n still in stdin.

The issue i believe is that you need to be getting length from a Pointer to the array not the array itself.
Try this, this code worked for me.
int ArrayLength(char* stringArray)
{
return strlen(stringArray);
}

Related

How do I request characters from a user, and then print the size of the characters?

I'm very new to C, any help would be greatly appreciated.
I can't use the <string.h> or <ctype.h> libraries.
This is the code I have:
int main(void)
{
char character;
printf("Introduce characters: ");
scanf(" %c", &character);
printf("\nSize of character: %d", sizeof(character)/sizeof(char));
return 0;
}
This only prints 1 as the size.
I read in another post that the problem was that initializing character by char character; would only let me store 1 single character. So, I modified it to be an array:
int main(void)
{
char character[10];
printf("Introduce maximum 10 characters: ");
scanf(" %s", character);
printf("\nSize of character: %d", sizeof(character)/sizeof(char));
return 0;
}
The problem now is that by doing character[10], it prints out that the size is 10. How would I go about fixing this?
sizeof(character)/sizeof(char) gives you the size of the array you declared, not the size of what the user has entered.
sizeof(character) gives the size of the entire array in bytes
sizeof(char) gives the size of a single character in bytes
So, when you do sizeof(character)/sizeof(char), you get the actual size (i.e. number of elements) of your array. What you are trying to achieve can be done with strlen(). But since you can't use <string.h>, you can write it yourself:
int strlen2(char *s)
{
int size;
for (size = 0; s[size]; size++)
;
return size;
}
Then use it like:
#include <stdio.h>
int main(void)
{
char character[10];
printf("Introduce maximum 10 characters: ");
scanf("%s", character);
printf("\nSize of character: %d", strlen2(character));
}
strlen2() counts the number of characters of your string, it stops counting when it encounters the first \0 character (null terminator).
Avoid using scanf() to read input
Your code is prone to bugs. If the user enters a string more than 9 characters long (don't forget the \0 is added at the end of your string), you'll get a buffer overflow, because character is only supposed to contain 10 characters. You would want to limit the number of characters read into your string:
scanf("%9s", character); // Read only the first 9 characters and ignore the rest
Moreover, scanf() is used to parse input, not to actually read it. Use fgets() instead:
#include <stdio.h>
#include <string.h> // for strcspn()
int main(void)
{
char character[10];
printf("Introduce maximum 10 characters: ");
if(!fgets(character, 10, stdin)) {
fprintf(stderr, "Error reading input.\n");
return 1;
}
character[strcspn(character, "\n")] = '\0'; // fgets() reads also `\n` so make sure to null-terminate the string
printf("\nSize of character: %zu", strlen(character));
}
fgets() accepts three arguments:
The first one is the array in which you want to store user input
The second one is the size of your array
The third one is the file stream you want to read from
It returns NULL on failure so you should check that as well.
Well if you can't use any headers, maybe you can create a custom strlen() function.
strlen() pretty much counts all character until the '\0' character is found. '\0' is used to signify the end of string and is automatically appended by scanf("%s",...).
#include <stdio.h>
size_t ms_length(const char *s)
{
size_t i = 0;
for (; s[i] != '\0'; i++)
;
return i;
}
int main(void)
{
char *str = "hello";
printf("%zu\n", ms_length(str));
return 0;
}
And if you want to be pedantic, you might even want to check the return value of scanf(), for input errors and also apply a limit to the character to be read to avoid a buffer overflow.
if (scanf(" %9s", character) != 1) /* 9 characters + 1 reserved for \0 */
{
/* handle error */
return 1;
}

Program fails to receive second input

It's just a code to receive user inputs in C program, but fails to do so and accepts null space as input. I have tried fgets() as well and the same thing keeps happening. Please advice on how to fix.
#include <math.h>
#include <stdio.h>
//#include <string.h>
#define len 16
int main(void)
{
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n,i=0,j=0;
printf("enter the number of cards:");
n = getchar();
//scanf("%d",&n);
int c1[len][n],card[len][n];
char buf[len];
printf("Enter card number:");
gets(buf);
system("Pause");
return (0);
}
"...code to receive user inputs in c program, but fails to do so and accepts null space as input..."
The reasons your existing code has problems is covered well in the comments under your post.
Consider a different approach: Define the following:
char inBuf[80] = {0};//
int numCards = 0;//Pick variable names that are descriptive (n is not)
int cardNum = 0;
bool isnum;
Then use it in conjunction with printf() etc.
printf("enter the number of cards:");
if(fgets(inBuf, sizeof(inBuf), stdin))//will read more than just a single char, eg. "12345"
{
int len = strlen(inBuf);
isnum = true;
for(int i=0;i<len;i++)
{
if(!isdigit(inBuf[i]))
{
isnum = false;
break;
}
}
if(isnum)
{
numCards = atoi(inBuf);
}
else
{
printf("input is not a number\n"
}
}
printf("Enter card number:");
if(fgets(inBuf, sizeof(inBuf), stdin))
{
...
Repeat variations of these lines as needed to read input from stdin, with modifications to accommodate assignment statements based on user input i.e. an integer (this example is covered), a floating point number, a string (eg. a persons name)
Although there is more that you can do to improve this, it is conceptually viable for your stated purpose...

Where does the newline character come from in C?

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';
}

C: Scanning While not EOF Loop Unexpected Results

I know there are many questions on the same topic of scanf until EOF is reached, but here's a particular case I haven't seen. Suppose I want to make a C program where the user enters a single character, and the program prints back the character and the number of times the user has entered a character until they press CTRL+D (EOF)
This is what I have:
#include <stdio.h>
int main(){
char thing;
int i=0;
while(scanf("%c", &thing) != EOF){
printf("time:%d, char:%c\n",i,thing);
i++;
}
return 0;
}
However, the output is not as expected. It's the following:
f
time:0, char:f
time:1, char:
p
time:2, char:p
time:3, char:
m
time:4, char:m
time:5, char:
I'm not too sure why i is being incremented again, and why printf gets executed again. Perhaps I'm missing something.
Try
#include <stdio.h>
int main(){
char thing;
int i=0;
while(scanf("%c", &thing) != EOF){
if (thing!='\n') {
printf("time:%d, char:%c\n",i,thing);
i++;
}
}
return 0;
}
#user2965071
char ch;
scanf("%c",&ch);
With such a snippet one reads any ASCII character from the stream including new line, return, tab, or escape. Thus, inside the loop I would test the symbol read with one of the ctype-functions.
Something like this:
#include <stdio.h>
#include <ctype.h>
int main(){
char thing;
int i=0;
while(1 == scanf("%c", &thing)){
if (isalnum(thing)) {
printf("time:%d, char:%c\n",i,thing);
i++;
}
}
return 0;
}
As for me, I think it's not a good idea to check scanf for returning EOF. I would rather check for the number of good read arguments.

C delete chars from string

I have program that asks to enter a string (mystring) and a char (ch). Then it deletes all entered chars (ch) from the string (mystring). For example "abcabc" and char 'a' then the result shoud be "bcbc".
-When I use scanf the program works nicely if the string does not have spaces. If I enter "abc abc abc" It reads and processes only the first 3 letters (until space).
Then I was advised to use gets(mystr); because it can read all the stirng. But when I use gets the result is the same as the input string and nothing happens.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100
int main(int argc, char *argv[])
{
char mystr[N] ,result[N];
char ch;
int i,k;
k=0;
printf("enter string \n");
//gets(mystr);///////////////////////////
//scanf("%s",&mystr);///////////////////
printf("enter char \n");
scanf("%c",&ch);
scanf("%c",&ch);
for ( i = 0; i <= strlen(mystr); i++ )
{
if (mystr[i] != ch)
{
result[k]=mystr[i];
k++;
}
}
puts(result);
system("pause");
return 0;
}
scanf("%c",&ch);
scanf("%c",&ch);
That second scanf is your problem. It's picking up the new-line character that you enter after the letter you want to remove (and overwrites the previous value of ch).
Get rid of it.
Please note, as the man page says:
Never use gets(). Because it is impossible to tell without knowing the data in advance how many
characters gets() will read, and because gets() will continue to store characters past the end of
the buffer, it is extremely dangerous to use. It has been used to break computer security. Use
fgets() instead.
hmm - not sure what the problem is - use getstr, but not scanf for the string, and it works for me in visual studio
int main(int argc, char *argv[])
{
char mystr[N] ,result[N];
char ch;
int i,k;
k=0;
printf("enter string \n");
gets(mystr);///////////////////////////
//scanf("%s",&mystr);///////////////////
printf("enter char \n");
scanf("%c",&ch);
// scanf("%c",&ch);
for ( i = 0; i <= strlen(mystr); i++ )
{
if (mystr[i] != ch)
{
result[k]=mystr[i];
k++;
}
}
puts(result);
system("pause");
return 0;
}
Use this one:
char temp[2];
scanf("%1s",temp);
ch = temp[0];
and use gets
scanf when used with chars has some problems (it gets the "old" new line). Here we "cheat" a little and we use scanf to get a string that can have up to one character. A string of 1 character clearly needs a second character for the terminator, so an array of 2 characters.
Be aware that using a scanf for the character to search, you won't be able to insert the space character.
Note that gets is an "evil" function. You can easily do buffer overruns using it (it doesn't check that the buffer is big enough). The "right" way to do it is normally: fgets(mystr, N, stdin); (the "file" variant of gets has a maximum number of characters that can be read and will append a \0 at the end). Note that if you insert 150 characters in a fgets, 99 will go to your string (because you gave 100 of max size), 1x \0 will be appended and the other characters will remain in the buffer "ready" for the next scanf/gets/fgets... (to test it, reduce the buffer to a smaller value, like 5 characters, and do some tests)
You can use fgets() as suggested by xanatos with a small hack, so you can reliably handle return characters. Just change the '\n' to '\0' in the string obtained using fgets.
And in your program, you forgot to terminate the new string with a '\0'.
So here's the code you're looking for.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 100
int main(int argc,char **argv){
char string[N],str1[N];
char ch;
int i,k = 0;
fgets(string,N,stdin);
string[strlen(string)-1] = '\0';
scanf("%c",&ch);
printf("\n%s , %c",string,ch);
for (i=0;i<=strlen(string);i++)
if(string[i] != ch)
str1[k++] = string[i];
str1[k] = '\0';
printf("\n%s , %s\n",string,str1);
return 0;
}

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