In my c class today I was troubling with the scanf() command, we were just learning pointers and we got a question asking us to get an array, and print it reversed without using the [] for anything but declaring the (int) array. Of course it seems like a piece of cake but not when you accidentally write:
scanf("%d ", arr + i);
Did you notice the space after the %d? Sure did take me a while to figure it out but for some reason that makes loops go crazy, and I wanted you guys to help me (and my teachers) to figure out why does that happen. Example:
#include <stdio.h>
#define LEN 10
void arrayInput(int * arr, unsigned int len);
void arrayReverseOutput(int * arr, unsigned int len);
int main(void)
{
int arr[LEN] = { 0 };
arrayInput(arr, LEN);
arrayReverseOutput(arr, LEN);
system("pause");
return 0;
}
void arrayInput(int * arr, unsigned int len)
{
unsigned int i = 0;
printf("Enter 10 numbers: ");
for (i = 0; i < len; i++)
{
//printf("i = %d \n", i); see what happens when you use this line
scanf("%d ", arr + i);
}
}
void arrayReverseOutput(int * arr, unsigned int len)
{
int i = 0;
printf("The numbers in reverse order: ");
for (i = --len; i >= 0; i--)
{
printf("%d ", *(arr + i));
}
}
I'm really curios to see what's going on with that scanf... it's as if it requires 2 inputs at the first time it runs but somehow still manages to put the inputs in their right position in the array...
Thanks for taking your time to read this <3
A space in the format string tells scanf() to match zero or more whitespace characters, until the match fails. Spaces (' '), newlines('\n'), carriage returns ('\r'), and tabs ('\t') are among the whitespace characters. When a space occurs at the end of a format string, scanf() will try to match whitespace characters from the input until no match is found. But, scanf() can only return when a match fails, or end of file is reached. Thus, in the case of a statement like:
scanf("%d ", arr + i);
the call to scanf() will appear to hang, greedily waiting for more input from the user. Whenever the Enter key is pressed, a newline is sent and matched by scanf(), which is still waiting for a failing match. Or end of file. You can escape such a loop by signalling end of file from the keyboard with Ctrl-D on Linux, or Ctrl-C on Windows.
It is almost always a mistake to terminate a scanf() format string with a space. A newline ('\n') is also a whitespace character, and has the same effect when placed at the end of a format string.
Note that spaces can be used effectively in scanf() format strings. For example:
int retval = scanf(" %c %c", &c1, &c2);
Here, if a previous IO operation has left a newline in the input stream (a not uncommon occurrence), the leading whitespace directs scanf() to read and ignore it. The second space in the format string tells scanf() to expect zero or more whitespace characters between the input characters to be converted. This allows the user to input the characters with an intervening space. Without the added whitespace, if a user entered a b\n, c2 would end up holding the value for a space character, and the b would be left behind in the input stream for the next IO operation to pick up. Also note that scanf() returns the number of successful conversions, allowing the program to check whether the input is as expected. If retval in the above line is anything other than 2, something has gone wrong.
Related
im trying to create a program to reverse a string and if the string has uppercase i must change it to lowercase and vice versa but when i use scanf[^\n] it cant input anything. the input process only succeed in first input
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main() {
int ctr1, ctr2, loop1, loop2;
char char1[1010];
char upp[1010], low[1010];
int len;
scanf("%d", &loop1);
for (ctr1 = 0; ctr1 < loop1; ctr1++) {
scanf("%[^\n]", char1);
len = strlen(char1);
for(ctr2=0;ctr2<len;ctr2++){
upp[ctr2] = toupper(char1[ctr2]);
low[ctr2] = tolower(char1[ctr2]);
if (char1[ctr2] == low[ctr2]) {
char1[ctr2] = upp[ctr2];
}
else if (char1[ctr2] == upp[ctr2]) {
char1[ctr2] = low[ctr2];
}
}
printf("Case #%d: ", ctr1 + 1);
for (ctr2 = len - 1; ctr2 >= 0; ctr2--) {
printf("%c", char1[ctr2]);
}
printf("\n");
}
}
The line
scanf("%[^\n]",char1);
will read everything in the line except for the newline character at the end of the line. When you run this line of code again in the next loop iteration, it will read nothing, because the newline character is still in the input stream.
For this reason, you must also remove the newline character from the input stream.
One simple way of doing this would be to make an additional call to getchar after every scanf call.
Another way of doing this is to instruct scanf to remove all whitespace characters (which includes newline characters) before attempting to match the input, like this:
scanf(" %[^\n]",char1);
However, using the scanf format specifiers %s or %[] without setting a maximum limit is generally not recommended, because a buffer overflow is possible. Therefore, in your case, it would be safer to write
scanf(" %1009[^\n]",char1);
to limit the number of matched characters to 1009, so that the total number of bytes written is limited to 1010 (including the terminating null character). That way, you can be sure that you won't be writing to the array out of bounds.
However, generally I recommend that you use the function fgets instead of scanf, as that function's behavior is more intuitive, as it always reads a whole line at a time (assuming that the provided memory buffer is large enough to store the entire line), including the newline character at the end of the line.
I'm new to programming but I wanted to make a program that gets as input a number, (length) and then stores a series of a's and b's of said length. Finally it should output the numbers as the ascii numbers. (so 97 and 98)
I thought I should malloc a char array of the size length and then do a for-loop over it and print everything as an integer.
The problem is however that I get a value 10 as the value of the first letter.
Thanks a lot for any help!
int main()
{
int length;
scanf("%d", &length);
char *matrix = malloc((length + 1 ) * sizeof(char));
for (int i = 0; i < length; i++)
{
scanf("%c", &matrix[i]);
}
for (int i = 0; i < length; i++)
{
printf("\n%d", matrix[i]);
}
return 0;
}
When inputting 3 on the first line and aba on the next line, I get 10 97 98.
However I expected it to be 97 98 97. Why do I get a value of 10 in the first place of the array?
Use
scanf(" %c", &matrix[i]);
^^^^
instead of
scanf("%c", &matrix[i]);
^^
When the format starts with a blank all white spaces are skipped.
From the C Standard (7.21.6.2 The fscanf function)
5 A directive composed of white-space character(s) is executed by
reading input up to the first non-white-space character (which remains
unread), or until no more characters can be read.
10 is the ASCII code of the (white space) new line character '\n' that was present in the input buffer after you entered the length of the array.
The first scanf() with the format string %d leaves a newline in the input buffer.
What happens here, is that your terminal collects input one full line at a time, passing it to the program, and then the scanf() only reads the digits from the buffer, leaving the newline character there for the next scanf() to see. The same would happen if you entered 10 abc: the space, abc and the newline would be left there.
This mismatch is not something people usually expect, and it's one of the things that makes scanf() annoying. I would suggest using fgets() instead to first read a full line, matching what the terminal gives, and then parse the number from it with sscanf() or strtol() (or atoi()).
This cleans up the issue at the point where the first line is read, instead of passing it on to the next input function to handle. Otherwise all your input functions are tied together, if the next input would be for a whole line with possible white space, you'd need to know if you expect to clear a pre-existing newline or not. (You could also replace the later scanf("%c") with getchar(), not that that matters with buffering though.)
That said, the scanf("%c")/getchar() loop may still see newlines if you enter lines that don't have as many characters as the loop expects, so if you don't want to see them at all, filter them out.
So, something like this:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int length;
char linebuf[100];
fgets(linebuf, 100, stdin);
length = strtol(linebuf, NULL, 10);
char *matrix = malloc(length + 1);
for (int i = 0; i < length; i++)
{
matrix[i] = getchar();
}
for (int i = 0; i < length; i++)
{
printf("\n%d", matrix[i]);
}
printf("\n");
return 0;
}
(The obvious downside of fgets() is that you have to decide on a maximum length for the input line, allocate a buffer and call another function in addition to it.)
I'm trying to set up a code that counts the whole string and doesn't stop after the first space that it finds. How do I do that?
I tried this kind of code but it just counts the first word then goes to display the number of letters in that first word.
So far this is what I have tried.
int main(){
char get[100];
int i, space=0, len=0, tot;
scanf("%s", get);
for (i=0; get[i]!='\0'; i++)
{
if (get[i] == ' ')
space++;
else
len++;
}
tot = space + len;
printf("%i", tot);
}
And
int main(){
char get[100];
int len;
scanf("%s", &get);
len = strlen(get);
printf("%i", len);
}
But would still get the same answer as the first one.
I expected that if the
input: The fox is gorgeous.
output: 19
But all I get is
input: The fox is gorgeous.
output: 3
strlen already includes spaces, since it counts the length of the string up to the terminating NUL character (zero, '\0').
Your problem is that that the %s conversion of scanf stops reading when it encounters whitespace, so your string never included it in the first place (you can verify this easily by printing out the string). (You could fix it by using different scanf conversions, but in general it's easier to get things right by reading with fgets – it also forces you to specify the buffer size, fixing the potential buffer overflow in your current code.)
The Answer by Arkku is correct in its diagnose.
However, if you wish to use scanf, you could do this:
scanf("%99[^\n]", get);
The 99 tells scanf not to read more than 99 characters, so your get buffer won't overflow. The [^\n] tells scanf to read any character until it encounters the newline character (when you hit enter).
As Chux pointed out, the code still has 2 issues.
When using scanf, it is always a good idea to check its return value, which is the number of items it could read. Also, indeed the \n remains in the input buffer when using the above syntax. So, you could do this:
if(scanf("%99[^\n]", get) == 0){
get[0] = 0; //Put in a NUL terminator if scanf read nothing
}
getchar(); //Remove the newline character from the input buffer
I will take one example to explain the concept.
main()
{
char s[20], i;
scanf("%[^\n]", &s);
while(s[i] != '\0') {
i++;
}
printf("%d", i);
return 0;
}
i have used c language and u can loop through the ending the part of the string and you will get the length. here i have used "EDIT SET CONVESRION METHOD" to read string, you can also gets to read.
I need to take an input of 20 words entered by the user put those into a 2D array and print that out
my current code is
char array2[20][20];
int i;
for(i=0;i<20;i++)
{
printf("enter a word\n");
scanf(" %[^\n]",array2[i]);
}
for(i=0;i<colsize2;i++)
{
printf("\n");
for(j=0;j<rowsize2;j++)
{
printf("%c",array2[i][j]);
}
}
(I have no idea what %[^\n] is but it works better than %c or %s)
there are no compiler errors and the program will run but when it prints the array after all the words have been entered all I get is complete garbage
like
aȪ(M▒awn-US▒ e▒(<▒▒t/▒▒▒(5h▒tr:▒(
qh▒tdle__000
HW5.exe▒`wauld▒(▒&Oe,▒*a▒+a▒▒
so much so that it takes a bit of scrolling to get back to the start of my program
I do have more in this program that's not in my question but I'm 99% sure it wouldn't mess with what I have here but if you do want to see the rest just ask
I literally just started programming so I don't know diddly squat about it yet so if you could keep that in mind when you answer also this is for a school assignment so it doesn't need to be perfect it just has to get the job done
thanks to whoever answers this I've been grappling with this for hours
The format string
" %[^\n]"
^ note the leading space
means that scanf will first read and discard any number of leading whitespace characters and then match any sequence of characters which does contain a newline. scanf can potentially overrun the buffer it save the string into if the input string is too large for the buffer invoking undefined behaviour. The %s format specifier means scanf skips the leading whitespace characters and reads the input string till it encounters a whitespace at which point it appends a terminating null byte to the buffer it writes into and then returns.
Therefore, what you need is
char array2[20][20];
int i;
for(i = 0; i < 20; i++)
scanf("%19s", array2[i]);
for(i = 0; i < 20; i++)
printf("%s\n", array2[i]);
Initialize your array to 0:
char array2[20][20] = { 0 } ;
And then print the string not every character:
for(i=0;i<20;i++)
{
printf("%s",array2[i]);
printf("\n");
}
My little program below shall take 5 numbers from the user, store them into an array of integers and use a function to print them out. Sincerly it doesn't work and nothing is printed out. I can't find a mistake, so i would be glad about any advice. Thanks.
#include <stdio.h>
void printarray(int intarray[], int n)
{
int i;
for(i = 0; i < n; i ++)
{
printf("%d", intarray[i]);
}
}
int main ()
{
const int n = 5;
int temp = 0;
int i;
int intarray [n];
char check;
printf("Please type in your numbers!\n");
for(i = 0; i < n; i ++)
{
printf("");
scanf("%d", &temp);
intarray[i] = temp;
}
printf("Do you want to print them out? (yes/no): ");
scanf("%c", &check);
if (check == 'y')
printarray(intarray, n);
getchar();
getchar();
getchar();
getchar();
return 0;
}
Change your output in printarray() to read:
printf("%d\n", intarray[i]);
^^
That will add a newline after each number.
Normally, output written to the console in C is buffered until a complete line is output. Your printarray() function does not write any newlines, so the output is buffered until you do print one. However, you wait for input from the user before printing a newline.
Change to that:
char check[2];
And also that:
scanf("%s", check);
if (!strcmp(check,"y"))
printarray(intarray, n);
Hope that helped. Your scanf("%c", &check); failed. Instead of y you end up having NL (ASCII code 10), which means the if part fails.
I don't know if it a nice fix though. Maybe someone could give a better one. Keep in mind if you input something bigger (eg yess) you going to get a bit unlucky ;)
Aside from the suggestions about printing the \n character after your array (which are correct), you also have to be careful with your scanf that expects the "yes/no" answer. Muggen was the first one to notice this (see his answer).
You used a %c specified in your scanf. %c specifier in scanf does not skip whitespace, which means that this scanf will read whatever whitespace was left in the input buffer after you entered your array. You hit the "Enter" key after entering the array, which put a newline character into the input buffer. After that scanf("%c", &check) will immediately read that pending newline character instead of waiting for you to enter "yes" or "no". That's another reason your code does not print anything.
In order to fix your scanf, you have to force it to skip all whitespace characters before reading the actual answer. You can do that by scanf(" %c", &check). Note the extra space before %c. Space character in scanf format string forces it to skip all continuous whitespace beginning from the current reading position. Newline character happens to be whitespace, so it will be ignored by this scanf.
printf("%d", intarray[i]);
add new line after this