I'm trying to write a program which given a certain number of input will output the product of the listed input (only accounting inputs of 0-9 and ignoring other).
For example:
input:345 would result output: 60, or another example would be, input: 3t4 and output: 12
I have given it many attempts and this is what I'm stuck with :
#include <stdio.h>
main(){
int c,i;
c = getchar();
i = 1;
while (c!= '\n'){
if (c>=48 && c<=57){
i=c*i;
c=getchar();
}
}
printf("%d",i);
}
How can i do this ?
Couple of issues in your code.
After each time the program encounters a non-numeric character it doesn't read further from the input. It reads the same character. Hence c=getchar() should be outside the if block
The multiplication happens for the char variable c. It should be converted to the actual number and then be multiplied. In your code you are multiplying its ascii value. Hence (c-48)*i instead of c*i
Use i=(c-48)*i; instead of i = c*i. So, the changed program would be:
#include <stdio.h>
main(){
int c,i;
c = getchar();
i = 1;
while (c!= '$'){
// printf("%c\n", c);
if (c>=48 && c<=57){
i=(c-48)*i;
}
c=getchar();
}
printf("%d",i);
}
This will ensure that you are using the numeric value of the digit, 0 not as ascii code of 0 but as simple 0.
use c=getchar() outside the if block. This should work.
int i,c;
i = 1;
while (i){
c=getchar();
if(c == '\n')
{
break;
}
if (c < 48 || c > 57)
{
i = -1;
break;
}
i = i * (c-48);
}
if (i == -1)
{
printf("Error. Non-Number Entered\n\n");
}
else {
printf("%d\n\n",i);
}
Related
Code:
int main(void) {
int c;
c = getchar();
while (c != EOF) {
putchar(c);
c = getchar();
printf("%d", c);
}
}
I enter a character and that character is returned with putchar(c), however, if I print that character I get a code back? example:
0 = 48010
1 = 49110
2 = 50210
etc...
what are these numbers? is this where the character is stored in memory or something?
cheers
So I believe I figured out the problem by introducing newlines
int main(void) {
int c;
c = getchar();
while (c != EOF) {
putchar(c);
printf("\n");
c = getchar();
printf("%d\n", c);
}
}
When I enter 1 this returns the following in the terminal window.
49
1
10
What's happening is I'm returning the ASII character code for 1 - > 49
returning the value I entered with getchar()
and returning 10 which is the linefeed value i.e. the enter command.
Right? without the newlines it was just concatenating them all, making it confusing.
In the following code example from K&R's book, if I replace putchar(c) with printf("%c", c) the code works the same. But if I replace it with printf("%d", c) it gives gibberish output.
#include <stdio.h>
int main() {
int c;
c = getchar();
while (c != EOF) {
putchar(c);
c = getchar();
}
}
From here, I learned that the getchar() converts the stdin to an 8-bit character whose value ranges from 0 to 255.
Now I want to print the value of c using putchar(c) in one line and printf("%d", c) in another line. So I wrote the following code:
#include <stdio.h>
int main() {
int c, b;
c = getchar();
b = c;
while (c != EOF && c != 10) {
printf("%c",c);
c = getchar();
}
printf("\n");
while (b != EOF && b != 10) {
printf("%d\t",b);
b = getchar();
}
}
I used the condition c != 10 as the newline character is read as 10 by getchar(). I expected the code to work as
$ ./a.out
783
783
55 56 51
but the program terminates as
$ ./a.out
783
783
55
I understand that getchar() takes input from stdin and the variable b is not stdin. So how should I copy the variable c to b so that my program works as I expect it to?
The problem is that your code does not (and cannot, as it stands) 'remember' the inputs you gave in the first loop. So, after you have finished that loop, your second loop is wanting to read in the characters for b (after it has output the first value, which is remembered from the earlier b = c line).
So, after outputting 55 (the integer value of the character 7), it is waiting for further input.
Probably the easiest way to get the output that you're looking for is to have an array of input characters. Then, you can output the %c values as you read them (as before), then re-run the outputs using the %d format in a subsequent for loop.
Here is a demonstration that does what I think you're after:
#include <stdio.h>
#define MAXINS 20 // Set to the maximum number of input characters you want to allow
int main()
{
int c[MAXINS];
int i = 0, n = 0;
c[0] = getchar();
while (i < MAXINS && c[i] != EOF && c[i] != 10) {
printf("%c", c[i]);
c[++i] = getchar();
++n;
}
printf("\n");
for (i = 0; i < n; ++i) {
printf("%d\t", (int)(c[i]));
}
return 0;
}
Feel free to ask for further clarification and/or explanation.
EDIT: On the point in the your first paragraph, "But if I replace it with printf("%d", c) it gives gibberish output." Well, when I try the following code and give 783 and then hit return (which generates a newline) I get the expected 55565110 as the output:
int main()
{
int c;
c = getchar();
while (c != EOF) {
printf("%d", c);
c = getchar();
}
return 0;
}
This may look like gibberish, but it's just the same output as you 'expect' in your later code, but without the spaces and with the addition of the 10 for the newline.
You need to have every character stored, because once you read a char from stdin, it is not present in stdin anymore.
Since you want the newline character in the end as a part of the input, you should use fgets to take the input.
Say you are taking an input that could have a maximum of 100 characters.
#include <stdio.h>
int main(void) {
char c[100]; // A char array
fgets(c,100,stdin);
int x=0;
while (c[x] != 10 && c[x] != EOF)
printf("%c",c[x++]);
printf("\n");
x = 0;
while (c[x] != 10 && c[x] != EOF) // You can simply compare it with the newline character too.
printf("%d ",c[x++]);
printf("\n");
return 0;
}
There are many ways to do this. You can also read stdin character-by-character ans store it in an array. However, since you need to display the ASCII values of the characters in another line after displaying the characters themselves, you will have to store them in an array.
You are copying only the first input, to copy the whole string you need to store each input in a buffer and check if the string doesn't overflow that buffer on each iteration:
int main(void)
{
enum {size = 256};
char buffer[size];
size_t count = 0;
int c;
while ((c = getchar()) && (c != '\n') && (c != EOF))
{
printf("%c", c);
if (count < size)
{
buffer[count++] = (char)c;
}
}
printf("\n");
for (size_t iter = 0; iter < count; iter++)
{
printf("%d\t", buffer[iter]);
}
printf("\n");
}
If you don't want to limit the buffer to an arbitrary size then you need to change your approach to use dynamic memory (realloc or a linked list)
#include<stdio.h>
int main() {
char text[1000];
int ch;
int index = 0;
while ((ch = getchar()) != EOF) {
text[index] = ch;
index++;
}
text[index] = '\0';
int i =0;
int num_Count=0;
int lower_Count=0;
int upper_Count =0;
while(i < index) {
if((text[i]>='0') && (text[i]<='9')){
num_Count ++;
i++;
}
else if((text[i]>='A') && (text[i]<='Z')){
upper_Count++;
i++;
}
else if((text[i]>='a') && (text[i] <='z')){
lower_Count++;
i++;
}
else
i++;
}
printf("%d %d %d", num_Count, lower_Count, upper_Count);
return 0;
}
It is a program that outputs the number of lower case, upper case, and number when the sentence is inputted.
For example,
Hi
Name
100
Would output 3 4 2
I keep seeing a runtime error.
The (while) part seems to be wrong.. I do not know what's wrong.
I ran your code in my system and checked for the input: Hi Name 100. The output I got is 3 4 2 which is the expected output. I feel the only place where the code can run in an infinite loop is while reading the inputs. Try to use ctrl+ d for EOF or ctrl+ z for windows.
Rest every thing is fine.
EOF means End Of File. It is used when you read data from a file. I suggest put a character like newline ('\n').
Im totally new to programming, I picked up a C manual to learn on my own. I dont want to use an array as Im trying to practice with getchar(). I want to be able to output an error message if the user enters anything other than a digit or an alphabet. I am also trying to practice the C library function isalpha() and isdigit(). This is what I wrote so far, but my output is not quite right.
Input 1: "hello"
Expected output : "valid output"
Input 2: "hello5"
Expected output : "valid output"
Input 3: "hello!"
Expected output : "invalid output"
But my program returns "valid input" for all the three inputs above
Please help a newbie try to learn. I greatly appreciate it.
#include <stdio.h>
#include <ctype.h>
int main ()
{
char ch;
int len;
int valid;
printf("Enter a word: ");
for(length = 0; (ch = getchar()) != '\n'; len++)
{
if(isalpha(ch) || isdigit(ch))
{
valid = 1;
}
else
{
valid = 0;
}
}
printf("Input length: %d\n", len);
if (valid == 1)
{
printf("Valid\n");
}
if(valid == 0)
{
printf("\n");
}
return 0;
}
You were almost there. Few pitfalls:
1st there is a typo for your variable name “length” instead of “len”.
2nd As Mitchel0022 stated, your program will always display “valid” providing the last character entered is valid because you reassign a new value for the variable ‘valid’ on each iteration. but you don’t have to use a ‘break statement since you need the loop to continue in order to get your length, so stick with your flag.
Now your program should run fine. Copy and paste the code below:
#include <stdio.h>
#include <ctype.h>
int main ()
{
char ch;
int len;
//flag is set true
int valid = 1;
printf("Enter a word: \n");
for(len = 0; (ch = getchar()) != '\n'; len++)
{
//change flag if character is not number nor letter
if(!isalpha(ch) && !isdigit(ch))
{
valid = 0;
}
}
printf("Input length: %d\n", len);
if (valid == 1)
{
printf("Valid\n");
}
else
{
printf("Invalid\n");
}
return 0;
}
The error here is when checking every character, by toggling valid to 0 and 1 really you are only looking at the last character. What you want to do is for every character as soon as you find one character that is not a digit/alpha character then exit
you can replace your if/else with something like this
vlaid = 1; //always assume its valid
for(int length = 0; (ch = getchar()) != '\n'; length++)
{
if(!isalnum(ch)) //if not a valid character
{
valid = 0; // string is not valid anymore
break; // we can exit the loop
}
}
My task is:
Write a program that reads input up to # and reports the number of times that the sequence ei occurs.
I wrote something that in most of the times works, but there are inputs when it dosent...
Like this input:(suppose to return 1)
sdlksldksdlskd
sdlsklsdks
sldklsdkeisldksdlk
#
number of combination is: 0
This is the code:
int main(void)
{
int index = 0;
int combinationTimes = 0;
int total = 0;
char userInput;
char wordChar[index];
printf("please enter your input:\n");
while ((userInput = getchar()) != '#')
{
if (userInput == '\n')
continue;
wordChar[index] = userInput;
index++;
total++;
}
for (index = 1; index < total; index++)
{
if (wordChar[index] == 'i')
{
if (wordChar[--index] == 'e')
{
combinationTimes++;
++index;
}
}
}
printf("number of combination is: %d", combinationTimes);
return 0;
}
Can you please tell me what am I not getting 1 using this input?
in the book he said to test it with "Receive your eieio award" and it worked...but after i played with it a little i see that not always.
It really doesn't seem necessary to read the file into an array. You just need to keep track of how many times ei is found before you read a # or reach EOF:
#include <stdio.h>
int main(void)
{
int c;
int ei_count = 0;
while ((c = getchar()) != EOF && c != '#')
{
if (c == 'e')
{
int c1 = getchar();
if (c1 == 'i')
ei_count++;
else if (c1 != EOF)
ungetc(c1, stdin);
}
}
printf("ei appeared %d times\n", ei_count);
return(0);
}
Testing (the program is called ei and is built from ei.c):
$ ei < ei.c
ei appeared 0 times
$ sed 1d ei.c | ei
ei appeared 1 times
$ sed 's/#/#/' ei.c | ei
ei appeared 4 times
$
The first one stops at the #include line, the second stops at the # in the comparison, and the third reads the entire file. It also gives the correct output for the sample data.
Analysing the code
Your primary problem is that you do not allocate any space for the array. Change the dimension of the array from index to, say, 4096. That'll be big enough for your testing purposes (but really the program should pay attention to the array and not overflowing it — but then I don't think the array is necessary at all; see the code above).
The next primary problem is that despite its name, getchar() returns an int, not a char. It can return any valid character plus a distinct value, EOF. So it must return a value that's bigger than a char. (One of two things happens if you use char. If char is a signed type, some valid character — often ÿ, y-umlaut, U+00FF, LATIN SMALL LETTER Y WITH DIAERESIS — is also treated as EOF even though it is just a character. If char is an unsigned type, then no input matches EOF. Neither is correct behaviour.)
Fixing that is easy, but your code does not detect EOF. Always handle EOF; the data may be malformatted. That's a simple fix in the code.
A tertiary problem is that the printf() statement does not end with a newline; it should.
Your test condition here is odd:
if (wordChar[--index] == 'e')
{
combinationTimes++;
++index;
}
It's odd to use one pre-increment and one post-increment, but that's just a consistency issue.
Worse, though, is what happens when the character i appears in the input and is not preceded by e. Consider the line #include <stdio.h>: you start with index as 1; that is an i, so you decrement index, but wordChar[0] is not an e, so you don't increment it again, but the end of the loop does, so the loop checks index 1 again, and keeps on going around the loop testing that the i is i and # is not e for a long time.
There's no reason to decrement and then increment index; just use:
if (wordChar[index-1] == 'e')
combinationTimes++;
With those fixed, your code behaves. You trouble was largely that you were using an array that was not big enough (being size 0), and you were overwriting quasi-random memory with the data you were reading.
#include <stdio.h>
int main(void)
{
int index = 0;
int combinationTimes = 0;
int total = 0;
int userInput;
char wordChar[4096];
printf("please enter your input:\n");
while ((userInput = getchar()) != '#' && userInput != EOF)
{
if (userInput == '\n')
continue;
wordChar[index] = userInput;
index++;
total++;
}
printf("total: %d\n", total);
for (index = 1; index < total; index++)
{
if (wordChar[index] == 'i')
{
if (wordChar[index-1] == 'e')
combinationTimes++;
}
}
printf("number of combination is: %d\n", combinationTimes);
return 0;
}
Note that you could reasonably write the nested if as:
if (wordChar[index] == 'i' && wordChar[index-1] == 'e')
combinationTimes++;
change your wordChar array value.
int main(void)
{
int index = 0;
int combinationTimes = 0;
int total = 0;
char userInput;
//char wordChar[index]; // index = 0
char wordChar[255]; // should change the value of array.
printf("please enter your input:\n");
while ((userInput = getchar()) != '#')
{
if (userInput == '\n')
continue;
wordChar[index] = userInput;
index++;
total++;
}
for (index = 1; index < total; index++)
{
if (wordChar[index] == 'i')
{
if (wordChar[--index] == 'e')
{
combinationTimes++;
++index;
}
}
}
printf("number of combination is: %d", combinationTimes);
return 0;
}
or maybe you can use pointer and then use malloc and realloc.