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I am beginner in c programming.I just want to know why this loop is not working properly.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
char x[8];
char t;
for (i = 0; i < 8; i++) {
scanf("%c", &t);
x[i] = t;
}
return 0;
}
Because when any input given from keyboard we need to press enter to confirm completion of input. This enter stay in buffer and if next input is char or string, stores enter in string or char var and do not wait to input that char or string. In this case, first input given at execution it stores char in X[0] and enter in x[1] and so on. So executes loop 8 time but it seems to 4 time because it ask input only four times. To check that put one printf in loop
it executes 8 times.
Whenever you press enter to submit, you are entering a whitespace character that is consuming one of your loop iterations.
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This is my code
#include <stdio.h>
main()
{
int c;
c=getchar();
while(c!=EOF)
{
int x;
x=(c!=EOF);
printf("%d",x);
putchar(c);
c=getchar();
}
}
Output (When I enter A):
A
1A1
Why is it not 1A only. Why is it repeating 1.Like first the program will take the value of c from getchar. Then it will go inside the loop. When condition will be true it will print the value of x and then value of c. Then again it should ask me for an input. Instead it is displaying another 1 then asking for the input. Kindly help.
It is because you entered A and <enter>. The <enter> produces the second 1.
You can press CTRL + D on unix or CTRL + Z on Windows to close the stdin of the program instead of pressing <enter>. This will give you the expected output.
With the <enter>: https://ideone.com/lWJ3Xz
and without: https://ideone.com/QsXiYz
a minimum debug effort and you will know yourself.
printf("FEOF=%d, C=0x%x\n",x,c);
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I want to enter value to my array and between 0 and 101 and trying to put zero or 101 will return an error but putting anything from 1-100 will allow me to enter a second number.
but i am stuck when write code to add a second number the loop keeps going in 1 and i cant figure out my error. any help would be apreciated
#include <stdio.h>
#include <stdlib.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
char letter;
int main()
{
int i;
int grades[12] ;
for ( i = 0; i < 12; i++)
do{//loops the following peice of code if values over 100 or less than 0 are entered
printf("Please Enter Grade 1:\n");
scanf("%f", & grades[i]);
if (grades > 100 || grades < 0)
printf("Invalid Entry please enter another between 0 and 100:\n");
}while(grades < 100 || grades > 0);
}
The mistake here is you're comparing the whole array to a value. What you want is one element, plus you need to test that you're outside of the acceptable bounds:
while(grades[i] > 100 || grades[i] < 0);
As Jonathan points out below, the logic in your code tests that it's within, which means only valid answers will loop forever, the opposite of what you wanted.
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I'm struggling to get Char to work. It keeps returning an error.
#include <stdio.h>
#include <cs50.h>
int main (void)
{
int tower_height;
char #;
// Inputs
do {
printf("Let's build! Give me a number between 0 and 23 'inclusive'.\n");
tower_height = GetInt();
}
while
(tower_height < 0 || tower_height > 23);
// Outputs
for (tower_height = 0; tower_height <= 23; tower_height++)
printf ("%c = tower_height - 2\n");
}
C identifier names may contain letters, underscore, and digits, as long
as the first character isn't a digit, and as long as the identifier
isn't a keyword. They may not contain #.
# is not a valid variable name.
As pointed out, # is not a valid variable name.
You can see how # is properly used in the first line of your code: #include <stdio.h>
Instead, call your char variable something that uses letters and numbers eg: char c;
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Please tell me, what is wrong here,it compiles but console crashes when i enter the number. I don't know what to write next, i will just write something to make my post possible.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int main(int argc, char *argv[])
{
unsigned int i,l,p,w;
printf("Enter natural number excluding 0: ");
scanf("%d",l);
p = 1;
for(i=1;i<=l;i++)
{
p=p*i;
}
w=p;
printf("\nFactorial of entered number %d",w);
return 0;
}
scanf("%d",l); you need to insert the address of l, which is &l.
You should also use %u for unsigned ints, not %d.
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#include <stdio.h>
int hledejznak(x)
{
int c;
int pocitadlo=0;
while((c=getchar())!=EOF)
{
if(x==c){
pocitadlo++;
while((c=getchar())!=32)
{
printf("%d\n",c);
};
};
};
return pocitadlo;
}
int main(int argc,char *argv[])
{
int znak=*argv[1];
printf("answer is %d",hledejznak(znak));
return 0;
}
Hi people, I need to count words containing character specified as argument at terminal
example: echo 'hello babe' | ./main e
Answer is 2
....because there are two words containing letter "e"
My code doesn't work, can you help me?
Thanks
Don't nest your loops; keep the outer one that processes each character read
Have a boolean variable initalized to false & set to true whenever you see the desired character.
Whenever a word ends, increment your counter if the flag is true. Either way, set the flag to false (to get ready for the next word). (Note that the last word may NOT end with a space.)
Only when you're processed all of the input should you print the value of the counter.