Let's say I have a string "abcd1234efgh". I want to split it into substrings of length 4, like:
abcd
1234
efgh
My C is rusty. Here's what I wrote:
#include<stdio.h>
#include<string.h>
int main(void){
int i,j;
char values[32]="abcd1234efgh";
char temp[10];
for(i=0;values[i]!='\0';){
for (j=0;j<4;j++,i++){
temp[i]=values[j];
printf("%c\n",values[j]);
}
printf("string temp:%s\n",temp);
}
return 0;
}
The output is obviously wrong because I'm not saving the index of the original string. Any tips on how to fix this? For strings with a length that is not a multiple of 4, I would like to pad the short substring with spaces.
This should do the trick if you're seeking to print only:
int len = strlen(values);
for (int off = 0; off < len; off += 4)
printf("%.4s\n", values+off);
If you want to do something else (as well) with the groups of 4, then I'd consider:
int len = strlen(values);
for (int off = 0; off < len; off += 4)
{
strncpy(temp, values+off, 4);
temp[4] = '\0';
…do as you will with temp…
}
NOTE : the code is to print in group of 4 and not break and store the string if size 4
if this is what you asked for
#include<stdio.h>
#include<string.h>
int main(void)
{
int i;
char values[32]="abcd1234efgh";
for(i=0;values[i]!='\0';)
{
if( i % 4 == 0 ) printf("\n");
printf("%c",values[i]);
}
return 0;
}
This should do the trick
#include <stdio.h>
#include <string.h>
int main() {
char *str = "abcd1234efgh";
size_t sub_len = 4;
size_t len = strlen(str);
size_t n = len / sub_len;
if(n * sub_len < len)
n += 1;
char temp[n][sub_len+1];
int i;
for (i = 0; i < n; ++i){
strncpy(temp[i], str + i*sub_len, sub_len);
temp[i][sub_len]='\0';
printf("string temp:%s\n", temp[i]);
}
return 0;
}
#include<stdio.h>
#include<string.h>
int main(void){
int i,j;
char values[32]="abcd1234efgh12";
char temp[10];
for(i=0;values[i]!='\0';){
for (j=0;j<4;j++,i++){
temp[j]=values[i];
}
while(j<4)
{
temp[j]=' ';
}
temp[j]='\0';
printf("string temp:%s\n",temp);
}
return 0;
}
Related
I am trying to convert int to binary as string but I can not.
Please help me. How to convert integer to binary, please tell me.
Input: 32
Output: 00100000
My code:
#include <stdio.h>
#include <string.h>
char converttobinary(int n)
{
int i;
int a[8];
char op;
for (i = 0; i < 8; i++)
{
a[i] = n % 2;
n = (n - a[i]) / 2;
}
for (i = 7; i >= 0; i--)
{
op = strcat(op, a[i]);
}
return op;
}
int main()
{
int n;
char str;
n = 254;
str = converttobinary(n);
printf("%c", str);
return 0;
}
I have tried to modify your solution with minimal changes to make it work. There are elegant solutions to convert Integer to Binary for example using shift operators.
One of the main issue in the code was you were using character instead of character array.
i.e char str; instead of char str[SIZE];
Also you were performing string operations on a single character. Additionally, iostream header file is for C++.
There is room for lot of improvements in the solution posted below (I only made your code work with minimal changes).
My suggestion is to make your C basics strong and approach this problem again.
#include <stdio.h>
#include <string.h>
void converttobinary(int n, char *op)
{
int i;
int a[8];
for (i = 0; i < 8; i++)
{
a[i] = n % 2;
n = (n - a[i]) / 2;
}
for (i = 7; i >= 0; i--)
{
op[i]=a[i];
}
}
int main()
{
int n,i;
char str[8];
n = 8;
converttobinary(n,str);
for (i = 7; i >= 0; i--)
{
printf(" %d ",str[i]);
}
return 0;
}
char *rev(char *str)
{
char *end = str + strlen(str) - 1;
char *saved = str;
while(end > str)
{
int tmp = *str;
*str++ = *end;
*end-- = tmp;
}
return saved;
}
char *tobin(char *buff, unsigned long long data)
{
char *saved = buff;
while(data)
{
*buff++ = (data & 1) + '0';
data >>= 1;
}
*buff = 0;
return rev(saved);
}
int main()
{
char x[128];
unsigned long long z = 0x103;
printf("%llu is 0b%s\n", z, tobin(x, z));
return 0;
}
I modify your code a little bit to make what you want,
the result of this code with
n = 10
is
00001010
In this code i shift the bits n positions of the imput and compare if there is 1 or 0 in this position and write a '1' if there is a 1 or a '0' if we have a 0.
#include <stdio.h>
void converttobinary(int n, char op[8]){
int auxiliar = n;
int i;
for (i = 0; i < 8; i++) {
auxiliar = auxiliar >> i;
if (auxiliar & 1 == 1){
op[7-i] = '1';
} else{
op[7-i] = '0';
}
auxiliar = n;
}
}
int main (void){
int n = 10;
int i;
char op[8];
converttobinary(n, op);
for(i = 7; i > -1; i--){
printf("%c",op[i]);
}
return 0;
}
I got given this assignment:
Write a C program that sequentially writes two strings into each other as shown in the figure below. Start with a string
consisting of “X”-es and with each iteration, the first and last X characters must be rewritten until the entire string is
rewritten and the final message is displayed.
Hint: Make use a function in the library, strlen(), to determine the length of a string.
It should output like this:
XXXXXXXXXXXXXXXXXXXXX
IXXXXXXXXXXXXXXXXXXX!
I XXXXXXXXXXXXXXXXXg!
I lXXXXXXXXXXXXXXXng!
I loXXXXXXXXXXXXXing!
I lovXXXXXXXXXXXming!
I loveXXXXXXXXXmming!
I love XXXXXXXamming!
I love CXXXXXramming!
I love C-XXXgramming!
I love C-PXogramming!
I love C-Programming!
Final String= I love C-Programming!
This is what I have so far:
#include <stdio.h>
#include <string.h>
int main(int argc, char **argv)
{
//data
char str[] = "I love C-Programming!";
int rows;
int columns;
int length = strlen(str);
int format =5;
//process
{
rows = 0;
while (rows <= length)
{
rows++;
}
while (rows > 0)
{
int count = length;
columns = rows - 1;
while (columns > 0)
{
printf("X");
columns--;
count --;
}
if (rows <= length)
{
printf("%.*s", count, str);
}
printf("\n");
rows-=2;
}
printf("%s", str);
}
//output
printf("\n");
printf("\n");
printf("Final String = %s\n", str);
return 0;
}
It doesn't display properly. Please help!
Thanks.
#include <stdio.h>
#include <string.h>
int main()
{
char s1[] = "XXXXXXXXXXXXXXXXXXXXX";
const char s2[] = "I love C-Programming!";
const int n = strlen(s1);
const int h = n / 2;
int i;
int j;
puts(s1);
for (i = 0, j = n - 1; i <= h; ++i, --j) {
s1[i] = s2[i];
s1[j] = s2[j];
puts(s1);
}
return 0;
}
Hi this is indeed a very simple program. Actually your teacher want you to write a program like below:-
int main(int argc, char **argv)
{
//data
char str1[] = "I love C-Programming!";
char str2[strlen(str1)];
memset(str2, 'X', sizeof(str2));//Set all the character to X
str2[strlen(str1)-1]=0;//end of string character value of '\0'
//int rows;
//int columns;
int length = strlen(str1);
//int format =5;
int i = 0;
int j = length - 1;
do
{
printf("%s\n", str2);//Print the second string first
str2[i]=str1[i];//copy from first character from str1
str2[j]=str1[j];//copy from last character from str1
//so in each iteration we are coping two characters from str1 to str2
}while(i++ != j-- );//once I and j are equal break the loop
printf("%s", str2);
/*
//process
{
rows = 0;
while (rows <= length)
{
rows++;
}
while (rows > 0)
{
int count = length;
columns = rows - 1;
while (columns > 0)
{
printf("X");
columns--;
count --;
}
if (rows <= length)
{
printf("%.*s", count, str);
}
printf("\n");
rows-=2;
}
printf("%s", str);
}
*/
//output
printf("\n");
printf("\n");
printf("Final String = %s\n", str2);
return 0;
}
It will out put like below:-
XXXXXXXXXXXXXXXXXXXX
IXXXXXXXXXXXXXXXXXXX!
I XXXXXXXXXXXXXXXXXg!
I lXXXXXXXXXXXXXXXng!
I loXXXXXXXXXXXXXing!
I lovXXXXXXXXXXXming!
I loveXXXXXXXXXmming!
I love XXXXXXXamming!
I love CXXXXXramming!
I love C-XXXgramming!
I love C-PXogramming!
I love C-Programming!
Final String = I love C-Programming!
I have the following code:
#include "stdafx.h"
#include "string.h"
#include "ctype.h"
/*selection sort*/
void swap(int A[], int j, int k)
{
int p = A[k];
int i;
for (i = 0; i < (k - j); i++)
{
A[k - i] = A[k - i - 1];
}
A[j] = p;
}
/*greatest number in an array*/
int max(int A[], int N, int k)
{
int max = k, i;
for (i = k; i < N; i++)
{
if (A[max] < A[i])
max = i;
}
return max;
}
int count_nonspace(const char* str)
{
int count = 0;
while(*str)
{
if(!isspace(*str++))
count++;
}
return count;
}
int _tmain(int argc, _TCHAR* argv[])
{
int a[256];
int i = 0, j = 0, count[256] = { 0 };
char string[100] = "Hello world";
for (i = 0; i < 100; i++)
{
for (j = 0; j<256; j++)
{
if (tolower(string[i]) == (j))
{
count[j]++;
}
}
}
for (j = 0; j<256; j++)
{
printf("\n%c -> %d \n", j, count[j]);
}
}
Program is calculating the number of apperances of each character in a string. Now it prints the number of apperances of all 256 characters, whereas i want it to prinf only the character with greatest number of apperances in a string. My idea was to use the selection sort method to the array with the nubmer of apperances, but this is not working, thus my question is how to printf only the character with the greatest number of apperances in the string?
If anybody would have doubts, this is NOT my homework question.
EDIT: I've just noticed that this code printf apperances of characters begining with "j" why is that?
I started typing this before the others showed up, so I'll post it anyway. This is probably nearly the most efficient (increasing efficiency would add some clutter) way of getting an answer, but it doesn't include code to ignore spaces, count characters without regard to case, etc (easy modifications).
most_frequent(const char * str)
{
unsigned counts[256];
unsigned char * cur;
unsigned pos, max;
/* set all counts to zero */
memset(counts, 0, sizeof(counts));
/* count occurences of each character */
for (cur = (unsigned char *)str; *cur; ++cur)
++counts[*cur];
/* find most frequent character */
for (max = 0, pos = 1; pos < 256; ++pos)
if ( counts[pos] > counts[max] )
max = pos;
printf("Character %c occurs %u times.\n", max, counts[max]);
}
Create an array with your char as index.
Keep incrementing the value in the array based on the characters read.
Now get the max out of the array which gives you the most occurring char in your input.
Code will look like:
#include <stdio.h>
#include<string.h>
#include<stdlib.h>
int main(void) {
char buf[100];
int i=0,max =0,t=0;
int a[256];
memset(a,0,sizeof(a));
fgets(buf,100,stdin);
buf[strlen(buf)-1] = '\0';
while(buf[i] != '\0')
{
a[(int)buf[i]]++;
i++;
}
i=0;
for(i=0;i<256;i++)
{
if(a[i] > max)
{
max = a[i];
t = i;
}
}
printf("The most occurring character is %c: Times: %d",t,max);
return 0;
}
Here is a solution for that, based on your own solution, and using qsort().
#include <string.h>
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
struct Frequency
{
int character;
int count;
};
int compare(const void *const lhs, const void *const rhs)
{
return ((struct Frequency *)rhs)->count - ((struct Frequency *)lhs)->count;
}
int main(int argc, char* argv[])
{
int i = 0, j = 0;
struct Frequency count[256];
memset(&count, 0, sizeof(count));
char string[100] = "Hello world";
for (i = 0 ; i < 100 ; i++)
{
for (j = 0 ; j < 256 ; j++)
{
count[j].character = j;
if (tolower(string[i]) == j)
{
count[j].count += 1;
}
}
}
qsort(count, sizeof(count) / sizeof(*count), sizeof(*count), compare);
/* skip the '\0' which was counted many times */
if (isprint(count[1].character))
printf("\nThe most popular character is: %c\n", count[1].character);
else
printf("\nThe most popular character is: \\%03x\n", count[1].character);
for (j = 0 ; j < 256 ; j++)
{
if (isprint(count[j].character))
printf("\n%c -> %d \n", count[j].character, count[j].count);
else
printf("\n\\%03x -> %d \n", count[j].character, count[j].count);
}
}
notice that the '\0' is set for all the remainig bytes in
char string[100] = "Hello world";
so the count of '\0' will be the highest.
You could use strlen() to skip '\0', in the counting loop, but don't
for (i = 0 ; i < strlen(string) ; ++i) ...
do it this way
size_t length = strlen(string);
for (i = 0 ; i < length ; ++i) ...
Lets say I have a string XYZ1-3.
I would like to convert it to a array of strings.
XYZ1,
XYZ2,
XYZ3.
is there an elegant way to do it in C?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char **expand(const char *string, int *num){//num : out var
char *id = strdup(string);
int start, end, len;
sscanf(string, "%*[A-Z]%n%d-%d", &len, &start, &end);
id[len] = '\0';
*num = end-start+1;
char **array = malloc(*num * sizeof(char*));
for(int i=0;i < *num ;++i){
len = snprintf(NULL, 0, "%s%d", id, start + i);
array[i] = malloc(++len);
sprintf(array[i], "%s%d", id, start + i);
}
free(id);
return array;
}
int main(){
int n;
char **array = expand("XYZ1-3", &n);
for(int i=0;i<n;++i){
printf("%s\n", array[i]);
free(array[i]);
}
free(array);
return 0;
}
Allow the non-alphabetical(not A-Z) id part
#include <ctype.h>
int id_length(const char *string){
//return length of id part.
int i, len;
for(i=0;string[i];++i);
if(i==0)return 0;
for(i=i-1;isdigit(string[i]) && i>=0;--i);
if(string[i]!='-') return 0;//bad format
for(i=i-1;isdigit(string[i]) && i>=0;--i);
return i+1;
}
char **expand(const char *string, int *num){//num : out var
char *id = strdup(string);
int start, end, len;
len = id_length(string);
sscanf(string+len, "%d-%d", &start, &end);
id[len] = '\0';
*num = end-start+1;
char **array = malloc(*num * sizeof(char*));
for(int i=0;i < *num ;++i){
len = snprintf(NULL, 0, "%s%d", id, start + i);
array[i] = malloc(++len);
sprintf(array[i], "%s%d", id, start + i);
}
free(id);
return array;
}
Try this--
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
int len,i,c,d,p,j;
char arr[50];
char arr2[50];
char arr3[30][30];
char temp[30];
scanf("%s",arr);
len=strlen(arr);//calculating length of entire input
for(i=0;i<len;i++)
{
if(arr[i]!='-')
arr2[i]=arr[i];//arr2[] will hold the string without the numeral
else
break;
}
c=(int)arr[i-1]-48;//char is converted into int
d=(int)arr[i+1]-48;
for(i=0;i<len-3;i++)
temp[i]=arr2[i];
p=0;
for(i=c;i<=d;i++)
{
temp[len-3]=(char)(i+48);//int is converted into character
for(j=0;j<=len-3;j++)
arr3[p][j]=temp[j];//this 2d array holds array of strings
p++;
}
for(i=0;i<=(d-c);i++)
{
for(j=0;j<=len-3;j++)
{
printf("%c",arr3[i][j]);//printing the strings one by one
}
printf("\n");
}
getch();
}
I know that getline is C++ standard but I need to read a line of digits:
123856
and save it to an array. But how to do this without spaces between given (as input) digits? I want a user input to be:
123856 (with no spaces) and then save it to an array (n element array) and after that, I want my array to look like this:
array[0] = 1;
array[1] = 2;
array[2] = 3;
array[3] = 8;
array[4] = 5;
array[5] = 6;
But how to make it in C, without a getline?
This is NOT what I want:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdbool.h>
int main(int argc, char **argv)
{
int t[4];
int i;
for(i=0; i<4; i++)
scanf("%d", &t[i]);
for(i=0; i<4; i++)
printf("%d\n", t[i]);
return 0;
}
If I understood you correct, the following should do it:
read the whole line
loop through the string as long as you get digits or the string ends
for every digit, place it's value in your array and increase the index by 1
while( ( c = getchar()) != EOF && c != '\n' && i < max ) {
/* If desired, add check for value outside of 0-9 */
array[ i++ ] = c - '0';
...
}
char arr[] = "1234567";
int intarr[10];
int count = 0;
for (char* ptr = arr; *ptr; ptr++) {
intarr[count] = *ptr - '0';
count++;
}
try this
#include <stdio.h>
#include <string.h>
main (int argc, char *argv[])
{
FILE *f;
int i=0;
int j=0;
char output[100];
char* output1[100];
char string[100];
char delims1[] = " ";
char delims2[] = "*";
char* result = NULL;
char* result3 = NULL;
int num;
//for (j=0; j<2; j++)
//{
//printf("%s",delims9[6]);
//}
f = fopen("text.txt","r");
//
while( fgets(string,sizeof(string),f) )
{
result = strtok( string, delims1 );
while( result != NULL )
{
output1[i]=result;
printf("%s\n",output1[i]);
result = strtok( NULL, delims1 );
i++;
}
for (num = 0; num < 100; i++ ) //
{ // Error On this array
printf("%s\n", output1[i]); //
} //
}
printf("\n%d",i/3+1);
return 0 ;
}
Ok, without using any string.
int digits = 123856;
int numofdigits = 1 + floor(log10(digits));
int digits_arr[numofdigits];
int i;
for(i = numofdigits-1; i >= 0; i--) {
digits_arr[i] = (int)floor(digits / pow(10, i)) % 10;
}
Try the below link... Same question asked here and get solution....
convert an integer number into an array
char * convertNumberIntoArray(unsigned int number) {
unsigned int length = (int)(log10((float)number)) + 1;
char * arr = (char *) malloc(length * sizeof(char)), * curr = arr;
do {
*curr++ = number % 10;
number /= 10;
} while (number != 0);
return arr;
}