I'm trying to do loop menu with some basic functions, everything is working fine apart of looping menu
in my opinion i have something wrong with while loop but i can't figure out what it is.
int main(void) {
char letter;
char status = 0;
printf ("--------------------------------------\n");
printf("a – Calculate the area of a rectangle\n");
printf("b – Calculate the area of a circle\n");
printf("c – Display a multiplication table\n");
printf("d – Add two numbers\n");
printf("x - exit program\n");
printf ("--------------------------------------\n");
scanf("%c",&letter);
while (status == 0)
{
if (letter == 'a' || letter == 'A')
{
}
if (letter == 'b'|| letter == 'B')
{
}
if(letter == 'c'|| letter == 'C')
{
}
if (letter == 'd'|| letter == 'D')
{
}
if(letter == 'x' || letter == 'X')
{
printf("shut down\n");
break;
}
status ++
}
return 0;
}
You need to pace a scanf(" %c",&letter); inside the loop body; otherwise, you will not get a chance to ever enter an x...
Please note the space before the %c, i.e. the " %c"-format, which captures any new line in the input buffer from a previous input.
Maybe you meant that status will be int and not char?
You should read input also in the beginning of loop, otherwise you will take only one input
After the first iteration, you advance status so it will exit the loop. Is this what you tried to achieve? I guess you meant for:
if(letter == 'x' || letter == 'X')
{
printf("shut down\n");
status ++
break;
}
Your loop will run only one time cause 'status ++' will work no matter the condition you should use it inside the x case -
if(letter == 'x' || letter == 'X')
{
printf("shut down\n");
status ++;
}
This should break your loop only after 'x' is entered.
Related
so I have been working on my assignment and I can't figure out how to put if...else statement in my do..while loop because want I run the program it doesn't loop. the output is like this.
[this the output that I got][1]
#include <stdio.h>
#include <ctype.h>
int main(void){
char alphabet;
char confirm;
int lowercase_vowel, uppercase_vowel;
int a =1;
do{
printf("Enter an alphabet: ");
scanf("%c", &alphabet);
// evaluates to 1 if variable c is a lowercase vowel
lowercase_vowel = (alphabet == 'a' || alphabet == 'e' || alphabet == 'i' || alphabet == 'o' || alphabet == 'u');
// evaluates to 1 if variable c is a uppercase vowel
uppercase_vowel = (alphabet == 'A' || alphabet == 'E' || alphabet == 'I' || alphabet == 'O' || alphabet == 'U');
if(!isalpha(alphabet))
{
printf("Error! Non-alphabetic character.\n");
}
else if(lowercase_vowel || uppercase_vowel)
{
printf("%c is a vowel.\n", alphabet);
}
else
{
printf("%c is a consonant.\n", alphabet);
}
printf("\nif u want to proceed enter 1, if not enter 0\n");
scanf("%d", &a);
}while( a == 1);
}
``
[1]: https://i.stack.imgur.com/oRK1G.png
do
{
// code
}while (a = 1);
This will create an infinite loop, because it will assign 1 to a, and because a is now a nonzero value, the condition is true, and it will loop again:
Maybe what you want is:
do
{
// code
}while (a == 1); // comparison, not assignment
Also:
scanf("%d", a);
should be:
scanf("%d", &a);
Brand new C coder here. In my first C course in school. I have experience in java but this course is all in C. I have homework to create a program that reads the contents of a file and counts the number of upper and lower case letters, vowels, consonants and digits. The program is not supposed to have any arguments, but will take a .txt file from the command line via redirection. My question is, how do I correct my current code to read from stdin each character of the file, whether it be a letter or a number? I'm really struggling with how read the contents of the file from stdin, read each character and then decide which category it belongs in. Any help would be appreciated. Thanks.
I'll be running the program like this...
$ program < testFile.txt
Where testFile.txt will contain the following text:
abcdefghijklmnopqrstuvwxyz
ABCDEFGHIJKLMNOPQRSTUVWXYZ
0123456789
int upper = 0; // Number of upper case letters
int lower = 0; // Number of lower case letters
int vowel = 0; // Number of vowels
int consonant = 0; // Number of constants
int digits = 0; // Number of digits
int total = 0; // Total number of characters in file
int i =0;
char value[100];
fgets(value, 100, stdin);
while(value[i] != '\0');
{
if (value[i] >= 'A' && value[i] <= 'Z')
{
upper++;
if (value[i] == 'A' || value[i] == 'E' || value[i] == 'I' || value[i] == 'O' || value[i] == 'U' || value[i] == 'Y')
{
vowel++;
}
else {
consonant++;
}
}
else if (value[i] >= 'a' && value[i] <= 'z')
{
lower++;
if (value[i] == 'a' || value[i] == 'e' || value[i] == 'i' || value[i] == 'o' || value[i] == 'u' || value[i] == 'y')
{
vowel++;
}
else {
consonant++;
}
}
else if (value[i] >= '0' && value[i] <= '9')
{
digits++;
}
total++;
i++;
}
printf("upper-case: %d", upper);
printf("\nlower-case: %d", lower);
printf("\nvowels: %d", vowel);
printf("\nconsonants: %d", consonant);
printf("\ndigits: %d", digits);
printf("\ntotal: %d", total);
printf("\n");
return 0;
I expect output to show how many upper case letters, lower case letters etc.
But once I run $ program < testFile.txt, it just sits there, no output to command line or anything.
Remove the semicolon after the while statement. :-)
while(value[i] != '\0');
This is your most obvious problem, it basically means:
while value[i] != '\0':
do nothing
end while
In other words, if it enters the loop, it will never exit it, because nothing changes that would affect the condition under which the loop continues.
There are other problems as well such as the fact that you will only process the first line rather than the whole file. The whole idea of using fgets and processing a line is unnecessary when you can just start with the following filter skeleton:
int ch;
while ((ch = getchar()) != EOF) {
/* process ch */
}
This will process an entire file character by character until all characters are done (or until an error occurs) so you can just tailor the body loop to do what you need - you've basically done that bit in your code with the loop over the line characters.
I would suggest not using the following code (since this is classwork) but you can also make better use of flow control constructs and library functions (from ctype.h and string.h), something like:
while ((ch = getchar()) != EOF) {
// Lib functions to detect upper/lower-case letters.
if (isupper(ch)) {
++upper;
} else if (islower(ch))
++lower;
}
// And to detect letter/digit type.
if (strchr("aeiouAEIOU", ch) != NULL) {
++vowel;
} else if (isalpha(ch)) {
++consonant;
} else if (isdigit(ch)) {
++digits;
}
++total;
}
This is particularly important since there's no actual guarantee that non-digit characters will be consecutive.
I'm facing a problem with what I enter with any unknown during the first time to the program. it will show me an infinite loop problem program closing. The program won't read the else statement.
char cont;
printf("Do u want continue\n");
scanf("%c", &cont);
getchar();
do
{
if (cont == 'y' || cont == 'Y')
{
selection();
}
else if (cont != 'n' || cont != 'N')
{
printf("Program Closing \n");
}
else
{
printf("Invalid Please Re-enter");
getchar();
scanf("%c", &cont);
}
} while (cont != 'n'&& cont != 'N');
let's dissect your code line by line starting with
scanf("%c", &cont);
This line would get a char value from stdin and put it into cont, which is a char so that's fine
getchar();
All I have to say for this is, why? it doesn't do anything useful, remove it.
Entering the loop now we have this statement
if (cont == 'y' || cont == 'Y')
this line is correct, it checks if the character is equal to y or Y
else if (cont != 'n' || cont != 'N')
this line is the main issue, your statement checks if cont is a value NOT equal to n or N, i.e. as a comment mentioned above, if the user put in the value a this line would return true, and then end the program. To correctly check if the user wants to exist you can use the same if statement used for y
if (cont == 'n' || cont == 'N')
if you replace the original if statement with this your program should work as expected. Just remember in the future that the != means not equal to, i.e. if the value is anything besides n or N return true. The == operator checks for equality as you saw above, so the line cont == 'n' means return true if cont is the same value as 'n'
printf("Invalid Please Re-enter");
getchar();
scanf("%c", &cont);
also as an extra note, please explain why you keep throwing in useless getchar()'s, those lines literally do nothing and you should remove them.
int main (void)
{
char tttarray[3][3] = {
{'1','2','3'},
{'4','5','6'},
{'7','8','9'}
};
int turn = 0,i,j,loc;
char XO;
while(turn<=9)
{
printf("Enter (x) or (o): ");
scanf("%c",&XO);
getchar();
printf("Enter number: ");
scanf("%d", &loc);
j = (loc - 1)%3;
i = (loc - 1)/3;
if(tttarray[i][j] == 'x' || 'o')
{
return(0);
printf("That spot has been taken!\n");
}
tttarray[i][j] = XO;
printArray(tttarray);
}
}
I'm only showing main the code for convenience. My problem is with this part:
if(tttarray[i][j] == 'x' || 'o')
{
return(0);
printf("That spot has been taken!\n");
}
This is because it always seems to be true which prevents the player from placing an 'x' or 'o' there. Am I missing something in the above code? How do prevent the user from going to a taken space?
You cannot chain the logical OR operator like you did. You need to change
if(tttarray[i][j] == 'x' || 'o')
to
if((tttarray[i][j] == 'x') || (tttarray[i][j] =='o'))
That said, a statement after a return statement does not make any sense. Control will never reach the printf() inside the if condition. Maybe what you want is to use
printf() statement
continue;
in this very order.
I've been trying to get this code to work but the loop does not seem to work? I am very new to C and I sort of get confused with the syntax of this language. However my loop is not functioning like how I want it to be. I want the if and else statement to work but no matter what input (right or wrong) it always outputs "thank you".
#include <stdio.h>
#include <stdlib.h>
int confirm()
{
char c;
printf("Confirm (y/n): ");
scanf("%c", &c);
while (scanf("%c", &c))
{
if (c = 'Y' && 'y' && 'N' && 'n')
{
printf("\nthank you");
break;
}
else
{
printf("\nInput not recognised, try again. \n");
printf("\nConfirm (y/n): ");
scanf("%c", &c);
}
}
}
int main(int argc, char* agrv[])
{
confirm();
return 0;
}
it won't ask to enter another output when the output is incorrect. It just keeps ending from the if statement, thus the loop is not running?
Please help.
There's nothing wrong with your loop - it's the if statement that's wrong.
This code compiles, but it does not do what you want it to do:
if (c = 'Y' && 'y' && 'N' && 'n')
= is an assignment; you need == to do a comparison
&& means "AND"; you need ||, which means an "OR"
You combine logical expressions, not constants with && or ||
The condition should be
if (c == 'Y' || c == 'y' || c == 'N' || c == 'n')
Also note that when you read single characters with %c, your program "sees" all characters, including whitespace. This is a problem, because the '\n' left over in the buffer will be passed to your program before Y or N. To fix this, add a space before %c to your format string:
scanf(" %c", &c)
// ^
// |
// Here
Your code also ignores the first character that it reads. I think this is not intentional, so remove the call of scanf before the loop. You should also remove the second scanf from the loop, leaving the only call to scanf in the loop header.
int confirm()
{
char c;
printf("Confirm (y/n): ");
//scanf("%c", &c);// <---------- needless
while (scanf("%c", &c)) //<----while loop will do `scanf("%c",&c)`, so previous line should be remove.
{
if (c == 'Y' || c == 'y' || c == 'N' || c == 'n')// <- &&(AND); ||(OR). Also, be careful that don't be lazy, [c == 'Y' || 'y' || 'N' || 'n'] can't to communicate with computer
{
printf("\nthank you");
break;
}
else
{
printf("\nInput not recognised, try again. \n");
printf("\nConfirm (y/n): ");
scanf("%c", &c);
}
}
}