An exercise asked to write a function that:
reads a sequence of alphabetic characters (without memorizing the sequence) that ends only when the users enters '\n'.
returns 1 if the number of capitalized letters went beyond the lower ones of at most an integer m, entered by the user, or 0 else.
I tried with the following code:
#include<stdio.h>
int read(int p,char c)
{
int M=0,m=0,d;
char A,Z,a,z;
while(c != '\n')
{
if(A<=c<=Z)
{
M++;
}
else if(a<=c<=z)
{
m++;
}
scanf("%c",&c);
}
if(M-m>0)
d=(m-M);
else
d=0;
if(d==0)
return 0;
else if (d<=p)
return 1;
}
int main()
{
int a,h;
char k;
scanf("%d", &h);
scanf("%c", &k);
a=read(h,k);
printf("%d\n",a);
return 0;
}
At this point, trying to execute the program with the gcc command, i noticed that the program was taking just the integer, let's say 2, and gave back 0 as if it entered in the function without taking the second scan on the character.
Besides the formal misconception and errors about the program and c function that i'm glad you rectify,
I was trying to understand, because as they say i'm trying to be self-taught, how scanf function and function work in general, when and to who priority is given.
For example in function read it's not clear to me when the value i'm returning to the function are taken putting a higher if as i did.
This isn't going to do what you probably expect
if(A<=c<=Z)
... for all sorts of reasons. Firstly, the values of A and Z are uninitialized. Second, the logic is written to be read by a mathematician, not a C compiler.
You almost certainly wanted this:
if('A'<=c && c<='Z')
... and remove the four variables char A,Z,a,z;
Note that use of character constants such as 'A' and 'Z' assumes a runtime environment using ASCII character sets. If you're interested in a more portable solution, you can look up isupper() and islower()
Related
So the function I declared doesn't seem to working as intended, and even so, I don't think that's the proper way to compare characters by incrementing the pointer, so I'm generally lost here. To be honest, pointers have always confused me and I really need to learn how to use them if I am going to get better at C. Thanks anyone for any help!
Here's the code I have, and if this helps, the purpose of the program is for you to enter a line of text, enter a single character to search in that line of text, and then find those characters using CharIsAt. (Will add the following later) the values stored in "found" will then be printed as well.
#include <stdio.h>
#define SIZE 41
int CharIsAt(char *pStr,char ch,int loc[],int mLoc);
int main(void){
char array[SIZE],search;
int found[SIZE],chars;
printf("Enter a line of text(empty line to quit): ");
while (fgets(array,SIZE, stdin)!=NULL && array[0]!='\n')
{
printf("Enter a character to search: ");
search=getchar();
chars=CharIsAt(array,search,found,SIZE);
}
return 0;
}
int CharIsAt(char *pStr,char ch,int loc[],int mLoc){
//Searches for ch in *pStr by incrementing a pointer to access
//and compare each character in *pStr to ch.
int i,x;
for (i=0;i<mLoc;i++){
if (strcmp(pStr[i],ch)==0){
//Stores index of ch's location to loc
loc[i]=pStr[i];
x++;
}
}
//Returns the number of times ch was found
return x;
}
EDIT: Flipped sign around in the for loop. Now the program gives me a "stopped working" error.
I'm trying to do an assignment for a class where we create a while loop using the scanf function. Basically the program is a calculator where the user should be able to type things like add 20 and my main function should be able to call up the functions from another .c then continue on in the loop.
However this is first time I am programming and I have no idea how to format or begin a loop that scans the input from the user, calls the appropriate mathematical function, applies it, then continues on in the loop.
Here's what I have done for the loop so far, which surprisingly, didn't work:
#include "calc.h"
#include "stdheader.h"
int main(int argc, char ** argv){
int c;
char token[81];
while(c != EOF){
if (scanf("%s", token) == "clear"){
calc.clear();
}else{
}
}
return 0;
}
and here is a sample of the functions that it should be able to call up (or what I have for them at least)
int local;
local = 0;
void clear(void){
local = 0;
printf("\n%d", local);
}
void add(int c){
local = local + c;
printf("\n%d", local);
}
In C, strings (and arrays more generally) are not first-class data types and cannot be compared by ==. Moreover scanf() does not return a string in any case - you should read the documentation carefully.
Change:
if (scanf("%s", token) == "clear")
to:
scanf( "%s", token ) ;
if( strcmp( token, "clear" )
Additionally the test c == EOF is dangerous because c is not initialised.
int c = 0 ;
My observation:
comparing string with == is not a suggested one. Instead try to use strcmp
scanf() returns only integer, you are comparing it to clear in if (scanf("%s", token) == "clear")
changing code like
if(scanf("%5s", token) == 1 && strcmp("clear",token) == 0)
Also what is the initial value of int c = ? without this how can you compare here while(c != EOF)
You need to look at the details of how scanf() works.
Here is a good reference.
scanf() returns the number of tokens read, an int not a string. You can't compare strings with == anyway. To compare strings you would have to use strcmp() from the string.h library.
You don't need to prepend calc on the front of your clear() function.
There are several errors here, you're best path forward might be to try something simpler first.
However this is my first time programming and I have no idea how to format or begin a loop that scans the input from the user, calls the appropriate mathematical function, applies it, then continues on in the loop.
When you want to solve a problem that is more complex than you think that you can solve it break it up in sub-problems. First design the user interface. What should the user enter to communicate with the program.
When you have a plan integrate the features that you need:
parsing user input
formatting of output
number processing (some arithmetic in your example).
A completely different approach to solve your problem.
Have array of function pointers.
typedef int (*func) (int,int);
func a[] = {add,sub,div,mul};
Have your UI like
printf("0 - add , 1- sub, 2 - div, 3 - mul\n");
scanf("%d",&op);
printf("Enter 2 numbers\n");
scanf("%d %d",&var1,&var2);
Now have
a[op](var1,var2);
Define your functions accordingly . For eg
int add(int x,int y)
{
return x+y ;
}
and so on.
I am previously a java programmer, but I'm now doing a C course at university (computer science major).
I need the user to be able to enter 3 chars,the first 2 being numbers, and the last 1 being either 'v' or 'h'.
For example "1 2 v".
I need the user to be able to enter it with the spaces in between each character.
This is my current code:
void manageInput(char box[][width]){
char move[4];
char input[16];
while(1){
scanf("%s", input);
int i = 0;
while(input[i] != 0){
if(input[i] != ' ' && input[i] != "\n"){
move[i] = input[i];
}
i++;
}
printf("%s\n", move);
makeMove(box, move);
printBox(box, height, width);
// TODO
if(move[0] == 'x'){
exit(0);
}
}
}
However if I run it, it works fine when I enter the chars with out spaces like "12v", but If I enter "1 2 v", it will print out "1", call printBox, then print out "2", then print out box again, and so on.
If someone could explain what I'm doing wrong here, I would appreciate it.
If someone could explain what I'm doing wrong here, I would appreciate it.
The short story is: Your code doesn't fulfill your requirements. It simply doesn't do what you want it to do.
Your requirements are:
All fields must be one character. This requirement isn't fulfilled by your code. Your code will mistakenly accept multiple characters per field.
There must be one space (exactly one space?) between the fields. This requirement isn't fulfilled by your code. There might be multiple spaces between the fields, and your code will mistakenly accept that.
In fact, your code invokes undefined behaviour by accessing the move array out of bounds. Consider that as a consequence of one of the above scenarios i might become some value higher than 3. What might happen in this code: move[i] = input[i];?
Your code is also way too complex. All of your functionality can be performed by scanf alone. It's a very powerful function, when you know how to use it correctly... I suggest reading and understanding the manual multiple times, when you have an opportunity. You'll learn a lot!
I notice something you neglected to mention from within the logic you have presented: It's expected that the first field might also be 'x', which corresponds to an exit usecase. This is a bad design; the caller has no opportunity to clean up... but I'll run with it. You really should use return (and return an int value or something, corresponding to error/success) instead.
Let us caste that last paragraph aside, because we can simply consider 'x' to be invalid input (and exit as a result), and I don't want to change the contracts of your functions; I'll leave that to you. The expression described so far appears to be int x = scanf("%1[0123456789]%*1[ ]%1[0123456789]%*1[ ]%1[vh]", a, b, c);.
Note that it is expected that a, b and c will have enough space to store a string of one byte in length. That is, their declaration should look like: char a[2], b[2], c[2];.
Make sure you check the return value (x, in the example)! If x is 3, it's safe to assume that the three variables a, b and c are safe to use. If x is 2, it's safe to assume that a and b are safe to use, and so on... If x is EOF or 0, none of them are safe to use.
By checking the return value, you can reject input that doesn't match that precise pattern, that is:
Fields that aren't exactly one byte in width will be rejected.
Too many or too few spaces will be rejected.
Something else popped up that you have neglected to mention, and it's also present within your code: Chux mentioned that you'll likely be expecting the input to be terminated with a '\n' (newline) character. This can also be implemented in a number of ways using scanf:
scanf("%1*[\n]"); will attempt to read and discard precisely one '\n' character, but there's no way to ensure that was successful. getchar would be more appropriate for that purpose; something along the lines of if (getchar() != '\n') { exit(EXIT_FAILURE); } might make sense, if you wish to ensure that the lines of input are perfectly formed and bomb out when they aren't... #define BOMB_OUT?
scanf("%*[^\n]"); scanf("%*c"); makes more sense; If you're interested in reading one item per line, then it makes sense to discard everything remaining on the line, and then the newline character itself. Note that your program should always tell the user when it's discarding or truncating input. You could also use getchar for this.
void manageInput(char box[][width]){
for (;;) {
char a[2], b[2], c[2];
int x = scanf("%1[0123456789]%*1[ ]%1[0123456789]%*1[ ]%1[vh]", a, b, c);
if (x != 3) {
/* INVALID INPUT should cause an error value to be returned!
* However, this function has no return value (which makes it
* poorly designed)... Calling `exit` gives no opportunity for
* calling code to clean up :(
*/
exit(EXIT_FAILURE);
}
if (getchar() != '\n') {
# ifdef BOMB_OUT
exit(EXIT_FAILURE);
# else
scanf("%*[^\n]");
getchar();
puts("NOTE: Excess input has been discarded.");
# endif
}
char move[4] = { a[0], b[0], c[0] };
printf("%s\n", move);
makeMove(box, move);
printBox(box, height, width);
// TODO
if(move[0] == 'x'){
exit(0);
}
}
}
%s reads a whitespace-delimited string with scanf, so if that's not what you want, it's not the thing to use. %c reads a single character, but does not skip whitespce, so you probably also want a (space) in your format to skip whitespace:
char input[3];
scanf(" %c %c %c", intput, input+1, input+2);
will read 3 non-whitespace characters and skip any whitespace before or between them. You should also check the return value of scanf to make sure that it is 3 -- if not, there was less than 3 characters in your input before an end-of-file was reached.
It's usuall a bad idea to read string via scanf because of potential buffer overflow. Consider using fscanf or better fgets as in
fgets(input, 15, stdin);
Note the extra byte for '\0'.
Also, you're comparing char to string here: input[i] != "\n". It should be input[i] != '\n' instead.
And btw you can just use something like
int x, y;
char d;
scanf("%d%d%c", &x, &y, &d);
This looks like two simple bugs.
You need to use separate indexes for move[] and input[]
int i = 0;
while(input[i] != 0){
if(input[i] != ' ' && input[i] != "\n"){
move[i] = input[i];
}
i++;
}
Imagine input of 1 2 v
input[0] != 0, so we enter the loop
it's not ' ' or '\n' either, so we copy input[0] to move[0]
so far so good
You increment i, and discover that input[1] == ' '
But then you increment i again
You discover that you are interested in input[2] (2) - so you copy it to move[2], rather than move[1]. Oops!
Then to make things worse, you never put an end-of-string character after the last valid character of move[].
I have written a small script to detect the full value from the user input with the getchar() function in C. As getchar() only returns the first character i tried to loop through it... The code I have tried myself is:
#include <stdio.h>
int main()
{
char a = getchar();
int b = strlen(a);
for(i=0; i<b; i++) {
printf("%c", a[i]);
}
return 0;
}
But this code does not give me the full value of the user input.
You can do looping part this way
int c;
while((c = getchar()) != '\n' && c != EOF)
{
printf("%c", c);
}
getchar() returns int, not char. And it only returns one char per iteration. It returns, however EOF once input terminates.
You do not check for EOF (you actually cannot detect that instantly when getchar() to char).
a is a char, not an array, neither a string, you cannot apply strlen() to it.
strlen() returns size_t, which is unsigned.
Enable most warnings, your compiler wants to help you.
Sidenote: char can be signed or unsigned.
Read a C book! Your code is soo broken and you confused multiple basic concepts. - no offense!
For a starter, try this one:
#include <stdio.h>
int main(void)
{
int ch;
while ( 1 ) {
ch = getchar();
x: if ( ch == EOF ) // done if input terminated
break;
printf("%c", ch); // %c takes an int-argument!
}
return 0;
}
If you want to terminate on other strings, too, #include <string.h> and replace line x: by:
if ( ch == EOF || strchr("\n\r\33", ch) )
That will terminate if ch is one of the chars listed in the string literal (here: newline, return, ESCape). However, it will also match ther terminating '\0' (not sure if you can enter that anyway).
Storing that into an array is shown in good C books (at least you will learn how to do it yourself).
Point 1: In your code, a is not of array type. you cannot use array subscript operator on that.
Point 2: In your code, strlen(a); is wrong. strlen() calculates the length of a string, i.e, a null terminated char array. You need to pass a pointer to a string to strlen().
Point 3: getchar() does not loop for itself. You need to put getchar() inside a loop to keep on reading the input.
Point 4: getchar() retruns an int. You should change the variable type accordingly.
Point 5: The recommended signature of main() is int main(void).
Keeping the above points in mind,we can write a pesudo-code, which will look something like
#include <stdio.h>
#define MAX 10
int main(void) // nice signature. :-)
{
char arr[MAX] = {0}; //to store the input
int ret = 0;
for(int i=0; i<MAX; i++) //don't want to overrrun array
{
if ( (ret = getchar())!= EOF) //yes, getchar() returns int
{
arr[i] = ret;
printf("%c", arr[i]);
}
else
;//error handling
}
return 0;
}
See here LIVE DEMO
getchar() : get a char (one character) not a string like you want
use fgets() : get a string or gets()(Not recommended) or scanf() (Not recommended)
but first you need to allocate the size of the string : char S[50]
or use a malloc ( #include<stdlib.h> ) :
char *S;
S=(char*)malloc(50);
It looks like you want to read a line (your question mentions a "full value" but you don't explain what that means).
You might simply use fgets for that purpose, with the limitation that you have to provide a fixed size line buffer (and handle - or ignore - the case when a line is larger than the buffer). So you would code
char linebuf[80];
memset (linebuf, 0, sizeof(linbuf)); // clear the buffer
char* lp = fgets(linebuf, sizeof(linebuf), stdin);
if (!lp) {
// handle end-of-file or error
}
else if (!strchr(lp, '\n')) {
/// too short linebuf
}
If you are on a POSIX system (e.g. Linux or MacOSX), you could use getline (which dynamically allocates a buffer). If you want some line edition facility on Linux, consider also readline(3)
Avoid as a plague the obsolete gets
Once you have read a line into some buffer, you can parse it (e.g. using manual parsing, or sscanf -notice the useful %n conversion specification, and test the result count of sscanf-, or strtol(3) -notice that it can give you the ending pointer- etc...).
I need to limit the input from a user to only positive values, and count the number of digits in that number. The user will only type in a (+/-) whole number up to 9 characters long.
I'm only allowed to use the scanf function and for, while, or do-while loops.(I saw in similar questions how to do this using getchar, but I can only use scanf). I'm not allowed to use arrays, or any other library besides stdio.h and math.h
I know that if I write:
n=scanf("%c%c%c%c%c",&a,&b,&c,&e,&f);
n will count the number of successful scanf conversions.
The problem i'm having is that when I define the input with char, it does everything I want except that the user MUST enter 5 characters. So if the user wants to input "55" he has to press "5" "5" "enter" "enter" "enter".
I need the program to move on after the first "enter" but also be flexible to receive a number up to 9 digits long.
again, I can't use getchar or anything fancy. Just the really basic stuff in C that you learn in the first 2 weeks.
Use scanf to read the number into a long int , then use a for loop with a /10 to count the number of digits
What do you want the program to do in case of a -ve number being entered?
#include<stdio.h>
int main()
{
long int a;
int b;
do
{
scanf ("%ld",&a);
if(a<0)
printf ("invalid input");
}while(a<0);
for(b=0;a!=0;b++,a=a/10);
printf("%d",b);
}
(does not handle -ve numbers specially)
Something like
#include <stdio.h>
int main(void)
{
char buffer[10] = { 0 };
size_t len;
scanf("%9[0-9]", buffer);
for(len = 0; buffer[len] != 0; len++) ;
printf("%zu '%s'\n", len, buffer);
return 0;
}
works, but I don't know if it fits your need.
EDIT (bits of explanation)
You can replace size_t with int (or unsigned int), though size_t is better. If you do, use %d or %u instead of %zu.
The basic idea is to exploit a feature of the format of scanf; the 9[0-9] says the input is a sequence of up to 9 char in the given set i.e. the digits from 0 to 9.
The for(...) is just a way to count char, a simple implementation of a strlen. Then we print the result.
The approach I would take would be the following.
Loops are allowed, so go ahead and set one up.
You need to have a variable somewhere that will keep track of what the current number is.
Think about typing out a number, one character at a time. What needs to happen to the current_number variable?
You need to stop the loop if a return key has been pressed.
Something like this should do for starters, but I'll leave the rest up to you, specifically what return_check(ch), update_state(current_val) and char_to_int(ch) looks like. Also note that rather than use a function, feel free to put your own function directly into the code.
int current_val=0;
int num_digits=0;
char ch="\0"
for (num_digits=0;return_check(ch) && num_digits<=9;num_digits++)
{
fscanf("%c");
current_val=update_state(current_val);
current_val=current_val+char_to_int(ch);
}
As for the logic in update_state(), think about what happens, one character at a time, if a user types in a number, like 123456789. How is current_val different from a 1 to a 12, and a 12 to a 123.
Can you wrap a loop around it, something like (I don't know if all of the syntax is right):
const int max_size=9
int n=0; //counter for number of chars entered
char a[max_size-1];
do {
scanf(%c,&a[n]);
n++;
} while (a[n] != '\r' && n<max_size)