I am trying to get 2, length of 8, inputs. With the code I have, my problem is that bit1[0] is null. Where is my problem ?
This is kind of debug I have, to see it null-ifies the first element
PS: Dont mind about the calculations I have at the bottom of the code. I am trying to create a calculator that does binary summation. I still couldn't try my code, it can be meaningless to do it in the way I do. But its not the question I am asking so please avoid giving advice on that one. Thank you.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char bit1[8];
char bit2[8];
int result[8];
int sum;
int dig1;
int dig2;
printf("Enter first binary number without any whitespaces!\t");
scanf("%s", bit1);
printf("Enter second binary number without any whitespaces!\t");
scanf(" %s", bit2);
int handle = 0;
puts(bit1);
puts(bit2);
for(int i = 7; i >= 0; i--){
//printf("%d", bit1[i] - '0');
dig1 = bit1[i] - '0';
dig2 = bit2[i] - '0';
if(dig1 + dig2 + handle < 2){
result[i] = dig1 + dig2 + handle;
handle = 0;
} else {
result[i] = (dig1 + dig2 + handle)%2;
handle = 1;
}
printf("dig1: %d\tdig2: %d\thandle: %d\n", dig1, dig2, handle);
printf("%d is added!\n", result[i]);
}
for(int i = 0; i < 8; i++){
printf("%c", result[i]);
}
printf(" And carry is: %d", handle);
return 0;
}
UPDATED
Working code is:
char bit1[9];
char bit2[9];
int result[8];
int sum;
int dig1;
int dig2;
printf("Enter first binary number without any whitespaces!\t");
scanf("%8s", bit1);
printf("Enter second binary number without any whitespaces!\t");
scanf("%8s", bit2);
Rest are the same
When you want to get a string (that is exactly 8 characters) from user input you should write:
char string[9];
scanf("%8s", string);
The reason you need char string[9] and not char string[8] is because a C-string always ends in the null-character (0 or '\0'). Thus, you need one extra space to store that null-character.
Don't forget to always use %8s. If the user input then exceeds the 8 characters you want, C will automatically cut off the exceeding characters.
Related
So, I was writing this code for counting the digit frequency i.e. the number of times the digits from 0-9 has appeared in a user inputted string(alphanumeric). So, I took the string, converted into integer and tried to store the frequency in "count" and print it but when I run the code, count is never getting incremented and the output comes all 0s. Would be grateful if anyone points out in which part my logic went wrong.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
// takes string input
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
//turns the string to int
int x = atoi(s);
int temp = x, len = 0;
//calculates string length
while (x != 0) {
x = x / 10;
len++;
}
x = temp;
//parses through the string and matches digits with each number
for (int j = 0; j < 10; j++){
int count = 0;
for(int i = 0; i < len; i++){
if(x % 10 == j){
count++;
}
x = x / 10;
}
x = temp;
printf("%d ", count);
}
return 0;
}
To write a correct and reasonable digit-counting program:
Do not allocate any buffer for this.
Create an array to count the number of times each digit occurs. The array should have ten elements, one for each digit.
Initialize the array to zero in each element.
In a loop, read one character at a time.
Leave the loop when the read routine (such as getchar) indicates end-of-file or a problem, or, if desired, returns a new-line or other character you wish to use as an end-of-input indication.
Inside the loop, check whether the character read is a digit. If the character read is a digit, increment the corresponding element of the array.
After the loop, execute a new loop to iterate through the digits.
Inside that loop, for each digit, print the count from the array element for that digit.
Your approach is way to complicated for a very easy task. This will do:
void numberOfDigits(const char *s, int hist[10]) {
while(*s) {
if(isdigit(*s))
hist[*s - '0']++;
s++;
}
}
It can be used like this:
int main(void) {
char buf[1024];
int hist[10];
fgets(buf, sizeof buf, stdin);
numberOfDigits(s, hist);
for(int i=0; i<10; i++)
printf("Digit %d occurs %d times\n", i, hist[i]);
}
This can also be quite easily achieved without a buffer if desired:
int ch;
int hist[10];
while((ch = getchar()) != EOF) {
if(isdigit(ch))
hist[ch - '0']++;
}
#include <stdio.h>
int main(void) {
int input = 1223330;
int freq[10] = {0};
input = abs(input);
while(input)
{
freq[input%10]++;
input /= 10;
}
for(int i=0; i<10; ++i)
{
printf("%d: %.*s\n", i, freq[i], "*************************************************");
}
return 0;
}
Output:
Success #stdin #stdout 0s 5668KB
0: *
1: *
2: **
3: ***
4:
5:
6:
7:
8:
9:
This app is currently limited by the size of an int (approximately 9 or 10 digits).
You can update it to use a long long easily, which will get you to about 19 digits.
Program to calculate the average of n numbers given by the user.
Okay so I have this program whose purpose is what you have read above. Its output is not quite right. I figured out what the problem is but couldn't find the solution as I am not a leet at programming (newbie actually). Here is the code:
#include <stdio.h>
int main(void) {
char user_data[100];
long int sum = 0;
double average;
unsigned int numbers_count = 0;
for (int i = 0; i <= 99; ++i)
user_data[i] = 0;
unsigned int numbers[100];
for (int i = 0; i <= 99; ++i)
numbers[i] = 0;
printf("Please enter the numbers:");
fgets(user_data, sizeof(user_data), stdin);
int i = 0;
while (user_data[i] != 0) {
sscanf(user_data, "%u", &numbers[i]);
++i;
}
i = 0;
while (numbers[i] != 0) {
sum += numbers[i];
++i;
}
i = 0;
while (numbers[i] != 0) {
++numbers_count;
++i;
}
average = (float)sum / (float)numbers_count;
printf("\n\nAverage of the entered numbers is: %f",average);
return 0;
}
Now here comes the problem.
When I enter an integer say 23, it gets stored into the user_data in two separate bytes. I added a loop to print the values of user_data[i] to figure out what was wrong.
i = 0;
while (i <= 99) {
printf("%c\n",user_data[i]);
++i;
}`
and the result was this
user_data insight
This was the first problem, here comes the second one.
I added another loop same like the above one to print the numbers stored in numbers[100] and figure out what was wrong and here is the output. Here's a sample
numbers stored in numbers[]
Now my main question is
How to extract the full number from user_data?
I believe it could be helpful to layout user_data after the fgets() of "23" (assuming Linux or Mac new line):
+-----+-----+----+----+
| '2' | '3' | \n | \0 | .....
+-----+-----+----+----+
0 1 2 3
Note that user_data[0] does not contain 2 (the number 2)! It contains '2' (the character '2') whose code is (again, assuming Linux) 0x32 (in hex or 50 in decimal).
This is why your attempt to print the values of user_data[] have not been fruitful: you were trying to print the representation of the number, not the number itself.
To convert that string to the integer it represents, you can do something like:
num = atoi(user_data)
The function atoi() does the work for you. A more flexible function is strtol() which does the same but for long int (and also can handle string that represents numbers in a base that is not 10).
I hope this answers to your question: How to extract the full number from user_data?
There are some other points where you should clean up and simplify your code, but you can open another question in case you need help.
Try this:
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int i;
char user_data[100];
long int sum = 0;
double average;
unsigned int numbers_count = 0;
for( i=0; i<= 99; ++i)
user_data[i] = 0;
unsigned int numbers[100];
for( i=0; i<= 99; ++i)
numbers[i] = 0;
printf("Please enter the numbers:");
fgets(user_data,sizeof(user_data),stdin);
//int p=0;//use with strtol(see further code)
i = 0;
int j;//this will store each number in numbers array.. so this is also the count of numbers stored - 1
for(j=0;;){
for(i=0;i<strlen(user_data);i++)
{
if(user_data[i]=='\n'){
break;
}
if(user_data[i]==' '){
j++;
i++;
//p=i;//to be used with strtol
}
numbers[j]=(numbers[j]*10)+((user_data[i]-48));//alternatively use => numbers[j]=strtol(user_data+p,NULL,10);
}
break;
}
i = 0;
while( i<=j)
{
sum += numbers[i];
++i;
}
average = (float)sum/(j+1);
printf("\n\nAverage of the entered numbers is: %f",average);
return 0;
}
Sample input
10 11 12
Sample output
11.00000000
I have shown two approaches to solve this:
One is straight-forward, subtract 48 from each char and add it to numbers array(ASCII manipulation) .
Other is to use strtol. Now strtol converts the number pointed by the char pointer(char array in this case) until the next char is not a number. So use pointer arithmetic to point to further numbers(like here I have added p(yeah I know p is not a good variable name, so does i and j!)).
There are more ways to solve like using atoi library functions.
regarding the posted code:
what happens if one of the numbers is zero?
What happens if the sum of the numbers exceeds the capacity of 'sum'
#include <stdio.h> // sscanf(), fgets(), printf()
#include <stdlib.h> // strtol()
#include <string.h> // strtok()
// eliminate the 'magic' number by giving it a meaningful name
#define MAX_INPUTS 100
int main(void)
{
// the array can be initialized upon declaration
// which eliminates the 'for()' loop to initialize it
// char user_data[100];
// and
// initialization not actually needed as
// the call to 'fgets()' will overlay the array
// and 'fgets()' always appends a NUL byte '\0'
char user_data[ MAX_INPUTS ];
long int sum = 0;
double average;
// following variable not needed
// unsigned int numbers_count = 0;
// following code block not needed when
// 'user_data[]' initialized at declaration
// for (int i = 0; i <= 99; ++i)
// user_data[i] = 0;
// not needed, see other comments
//unsigned int numbers[100];
// not needed, as 'numbers' is eliminated
// for (int i = 0; i <= 99; ++i)
// numbers[i] = 0;
printf("Please enter the numbers:");
// should be checking the returned value
// to assure it is not NULL
// And
// this call to 'fgets()' is expecting
// all the numbers to be on a single input line
// so that could be a problem
fgets(user_data, sizeof(user_data), stdin);
// the following two code blocks will not extract the numbers
// for a number of reasons including that 'sscanf()'
// does not advance through the 'user_data[]' array
// int i = 0;
// while (user_data[i] != 0) {
// sscanf(user_data, "%u", &numbers[i]);
// ++i;
// }
// i = 0;
// while (numbers[i] != 0) {
// sum += numbers[i];
// ++i;
// }
// suggest the following,
// which also eliminates the need for 'numbers[]'
// note: the literal " \n" has both a space and a newline
// because the user is expected to enter numbers,
// separated by a space and
// 'fgets()' also inputs the newline
int i = 0;
char *token = strtok( user_data, " \n");
while( token )
{
// not everyone likes 'atoi()'
// mostly because there is no indication of any error event
// suggest using: 'strtol()'
//sum += atoi( token );
sum += strtol( token, NULL, 10 ) // could add error checking
i++;
token = strtok( NULL, " \n" );
}
// the value assigned to 'numbers_count'
// is already available in 'i'
// suggest eliminate the following code block
// and
// eliminate the 'numbers_count' variable
// i = 0;
// while (numbers[i] != 0) {
// ++numbers_count;
// ++i;
// }
// 'average' is declared as a 'double',
// so the casting should be to 'double'
// and
// if incorporating the prior comment about 'numbers_count'
// average = (float)sum / (float)numbers_count;
average = (double)sum / (double)i'
// to have the text immediately displayed on the terminal
// place a '\n' at the end of the format string.
// without adding the '\n' the text only displays
// as the program exits
// printf("\n\nAverage of the entered numbers is: %f",average);
printf("\n\nAverage of the entered numbers is: %f\n",average);
return 0;
} // end function: main
Hi how would you count the number of occurences in the given word like shown below because with the program I have right now it doesn't seem to room correctly.
#include <stdio.h>
int main(void)
{
char a;
char lang[] = "pneumonoultramicroscopicsilicovolcanoconiosis";
char i = 0;
char count = 0;
printf("pneumonoultramicroscopicsilicovolcanoconiosis\n");
printf("\nEnter the letter you want to find the number of\n");
scanf("%c", &lang);
for (i = 0; i <= 46; i++)
if (a == lang[i]) {
count++;
}
printf("Number of %c is %d..\n", a, count);
return 0;
Your scanf is the problem.
Try:
scanf("%c",&a);
declare count as int instead of char
also change scanf to take input a
#include <stdio.h>
int main(void)
{
char a;
char lang[] = "pneumonoultramicroscopicsilicovolcanoconiosis";
char i = 0;
int count = 0;
printf("pneumonoultramicroscopicsilicovolcanoconiosis\n");
printf("\nEnter the letter you want to find the number of\n");
scanf("%c", &a);
for (i = 0; i <= 46; i++)
if (a == lang[i]) {
count++;
}
printf("Number of %c is %d..\n", a, count);
return 0;
}
You probably shouldn't hard code the max length of the string (46) if at all possible in case you are given a longer string, but assuming it's an assigned assignment that is set it shouldn't be a problem.
i and count should also be ints if possible for a bigger size count. And &lang should be &a since lang is already assigned while a is your checker.
I need to code a program that gets input values for a string, then it ignores the characters that are not digits and it uses the digits from the string to create an integer and display it. here are some strings turned into integers as stated in the exercise.
I wanted to go through the string as through a vector, then test if the each position is a digit using isdigit(s[i]), then put these values in another vector which creates a number using the digits. At the end it's supposed to output the number. I can't for the life of it figure what's wrong, please help.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main()
{
char *s;
scanf("%s", s);
printf("%s\n", s);
int i, n=0, v[100], nr=0;
for(i=0; i<strlen(s); i++)
{
if (isdigit(s[i]) == 1)
{
v[i] = s[i];
n++;
}
}
for(i=0;i<n;i++)
{
printf("%c\n", v[i]);
}
for(i=0; i<n; i++)
{
nr = nr * 10;
nr = nr + v[i];
}
printf("%d", nr);
return 0;
}
The pointer s is unintialized which is your major problem. But there are other problems too.
isdigit() is documented to return a non-zero return code which is not necessarily 1.
The argument to isdigit() needs to be cast to unsigned char to avoid potential undefined behaviour.
Your array v is also using the same index variable i - which is not right. Use a different variable to index v when you store the digits.
You need to subtract '0' to get the each digits integer equivalent.
scanf()'s format %s can't handle inputs with space (among other problems). So, use fgets().
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main(void)
{
char s[256];
fgets(s, sizeof s, stdin);
s[strcspn(s, "\n")] = 0; /* remove trailing newline if present */
printf("%s\n", s);
int i, n = 0, v[100], nr = 0;
size_t j = 0;
for(i = 0; i < s[i]; i++)
{
if (isdigit((unsigned char)s[i]))
{
v[j++] = s[i];
n++;
}
}
for(i = 0;i < j; i++)
{
printf("%c\n", v[i]);
}
if (j) { /* No digit was seen */
int multiply = 1;
for(i= j-1 ; i >= 0; i--) {
nr = nr + (v[i] - '0') * multiply;
multiply *= 10;
}
}
printf("%d", nr);
return 0;
}
In addition be aware of integer overflow of nr (and/or multiply) can't hold if your input contains too many digits.
Another potential source of issue is that if you input over 100 digits then it'll overflow the array v, leading to undefined behaviour.
Thanks a lot for your help, i followed someone's advice and replaced
v[i] = s[i] -> v[n] = s[i] and changed char *s with char s[100]
now it works perfectly, i got rid of the variable nr and just output the numbers without separating them through \n . Thanks for the debugger comment too, I didn't know I can use that effectively.
Firstly, you did not allocate any memory, I changed that to a fixed array.
Your use of scanf will stop at the first space (as in the first example input).
Next, you don't use the right array index when writing digits int v[]. However I have removed all that and simply used any digit that occurs.
You did not read the man page for isdigit. It does not return 1 for a digit. It returns a nonzero value so I removed the explicit test and left it as implicit for non-0 result.
I changed the string length and loop types to size_t, moving the multiple strlen calls ouside of the loop.
You have also not seen that digits' character values are not integer values, so I have subtracted '0' to make them so.
Lastly I changed the target type to unsigned since you will ignore any minus sign.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main(void)
{
char s[100]; // allocate memory
unsigned nr = 0; // you never check a `-` sign
size_t i, len; // correct types
if(fgets(s, sizeof s, stdin) != NULL) { // scanf stops at `space` in you example
len = strlen(s); // outside of the loop
for(i=0; i<len; i++) {
if(isdigit(s[i])) { // do not test specifically == 1
nr = nr * 10;
nr = nr + s[i] - '0'; // character adjustment
}
}
printf("%u\n", nr); // unsigned
}
return 0;
}
Program session:
a2c3 8*5+=
2385
Just use this
#include <stdio.h>
#include <ctype.h>
int main()
{
int c;
while ((c = getchar()) != EOF) {
switch (c) {
case '0': case '1': case '2': case '3': case '4':
case '5': case '6': case '7': case '8': case '9':
case '\n':
putchar(c); break;
}
}
return 0;
} /* main */
This is a sample execution:
$ pru_$$
aspj pjsd psajf pasdjfpaojfdapsjd 2 4 1
241
1089u 0u 309u1309u98u 093u82 40981u2 982u4 09832u4901u 409u 019u019u 0u3 0ue
10890309130998093824098129824098324901409019019030
Very elegant! :)
I want to scan in a string that can take at least 200 characters and then I want to convert the string to an int, so that I can print it with e.g. printf("%d", digit).
How can I write a function kinda like this I've written here
(this one does not work!):
int main()
{
char car[200];
int number;
int i,x;
int sum = 0;
printf("Write in number: \n");
scanf("%c", &car);
for (i=0; i<200; i++) {
if (car[i] != '\0') {
x = car[i]-'0';
sum = sum + x;
if (i != 0) {
sum = sum*10;
}
}
}
return 0;
}
first:
scanf("%c", &car);
from man scanf:
c
Matches a sequence of characters whose length is specified by the maximum field width (default 1); the next pointer
must be a pointer to char, and there must be enough
room for all the characters (no terminating null byte is added). The usual skip of leading white space is suppressed. To
skip white space first, use an explicit space
in the format.
So you're reading exactly one character, not a whole string.
Then, you rely on a terminating null byte being added, which isn't happening. Use %199s instead, leaving enough room for the terminating null byte.
Then, considering no int in this world should have enough space for numbers with 199 decimal digits, you should think about your 200 character buffer.
If your goal is not to write such a function for educational purposes, but because you need one:
int number;
scanf("%d", &number);
does exactly that: Read one number from the input, and place it in number.
scanf("%s", car);
You need to read a string %s not a single char %c. Also char array will decay to a pointer when passed as an argument, so you shouldn't use &char.
Here is the solution,Start from the end of the string because units place is from right to left then increment the units place to ten's place then 100 and so on.
#include<stdio.h>
#include<string.h>
int main()
{
char input[9];
int digit,number=0,i=1;
printf("Enter Number:\n");
scanf("%s",input);
digit=strlen(input)-1;
while(digit>=0)
{
number=number + i*(input[digit]-'0');
digit=digit-1;
i=i*10;
}
printf("%d",number);
return 0;
}
Close. Suggested changes below.
Use char car[200+1] to "take at least 200 characters".
Use "%200s" rather than "%c" to read a string rather than a char and to limit its input.
//int main()
int main(void) {
// char car[200];
char car[200+1];
int number;
int i,x;
int sum = 0;
printf("Write in number: \n");
// scanf("%c", &car);
scanf("%200s", car);
// for (i=0; i<200; i++) {
for (i=0; car[i]; i++) {
// if (car[i] != '\0'){
x = car[i]-'0';
// sum = sum + x;
sum = 10*sum + x;
// if (i != 0) { sum = sum*10; }
}
printf("%d\n", sum)
return 0;
}
Unless input is like "000000000000000000000000000123", 200 digits will certainly overflow sum.
To detect that
x = car[i]-'0';
if (sum >= INT_MAX/10 && (sum > INT_MAX/10 || x > INT_MAX%10)) {
x = INT_MAX;
// maybe set an error flag
break;
}
sum = 10*sum + x;
An int has an maximum value of INT_MAX from #include < limits.h>. It is at least 32767. Some platforms use 16-bit, 32-bit or 64-bit int Other ranges are possible. Let us assume a worst case of 128-bit. That would need about 39 decimal digits. Leaving room for a sign and terminating null character, suggest
char car[39+1+1];