This code doesnt seem to work for me. it outputs the number 17 which is obviously wrong.the counter should go up if it encounters numbers like 5, 15, 25, 50 ect.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main(void) {
int i;
int counter;
char num[4322];
for (i = 1; i < sizeof(num); i++){
num[i] = i;
if ( strstr(&num[i], "5")){
counter = counter + 1;
}
}
printf("%d", counter);
return 0;
}
You didn't convert the numbers to strings, and you passed what are not a null-terminated strings to strstr().
You also forgot to initialize counter.
You should check the digits one-by-one. Try this:
#include <stdio.h>
int main(void) {
int max = 4322;
int target = 5;
int i, cur, counter = 0;
for (i = 1; i <= max; i++) {
for (cur = i; cur > 0; cur /= 10) {
if (cur % 10 == target) counter++;
}
}
printf("%d\n", counter);
return 0;
}
There are many problems in this code.
char num[4322];
This is an array of 4322 characters. You probably didn't want that, I'm not sure why it is here.
num[i] = i;
This is invalid, because you cannot assign large numbers to a char. This will not convert the number to a string, this will convert the number to a char which is different. So 5 does not become "5" but it becomes \x05.
strstr(&num[i], "5")
This will search for "5", but since the array is not nul-terminated, it is incorrect. Also note that you are storing
int counter;
Note that counter is uninitialized. This is an error.
Here is a similar program, which is correct (but not "efficient" per se):
#include <stdio.h>
#include <string.h>
int main() {
int counter = 0; // initialize to 0
for (int i = 0; i < 4322; i++) {
// Put the number in a string
char buf[10];
snprintf(buf, sizeof(buf), "%d", i);
// Count 5s
for (char *p = buf; *p; p++) {
if (*p == '5') {
counter++;
}
}
}
printf("%d", counter);
}
Note that #MikeCAT's answer skips the int -> string conversion, which is somewhat unnecessary.
Related
I want to compare the integers in a string with integers (0-9) and I wrote this -
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char num[100];
int count = 0;
scanf("%s", num);
int len = strlen(num);
for (int i = 0; i <= 9; i++)
{
for (int j = 0; j <= len; j++)
{
if (i == (num[j] - '0'))
{
count++;
}
}
printf("%d ", count);
count = 0;
}
return 0;
}
No problems with this (works in most cases but it is failing in few cases). So can you please give me alternate and best idea to do this?
Thanks in advance
Complete pic -
The root cause is not in char comparison, but in the under-allocated buffer:
char num[100];
The assignment constraint is:
1 <= len(num) <= 1000
After increasing the buffer size, all the tests pass.
Besides a too small input buffer (i.e. 100 instead of 1001), I think your approach is too complex.
Instead of a nested loop, I'll suggest an array to count the frequency, i.e. an array with 10 elements so that you have a counter for each digit.
int main() {
char num[1001]; // 1000 chars + 1 zero termination
int count[10] = {0}; // Array of 10 zero initialized counters, one for each digit
scanf("%1000s", num); // At max accept 1000 chars input
char* p = num;
while (*p)
{
if (isdigit(*p) ++count[*p - '0'];
++p;
}
for (int i = 0; i < 10; ++i) printf("%d ", count[i]);
puts("");
return 0;
}
If you don't want to use isdigit you can instead do:
if (*p >= '0' && *p <= '9') ++count[*p - '0'];
In this code you "randomly" fill your array with letters and if it hits the word you are looking for returns the amount of trys to get it correct. What I want to implement is a way of checking if it is spelled correct from the first character going onwards. Any ideas?
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
int main() {
char *search= "SOMETHING";
int clk = clock();
printf ("clk=%d\n",clk);
srand(clk);
int len = strlen(search);
char *tafel = (char *) malloc (len+1);
tafel[len] = 0;
int pos;
char letter;
long i=0;
char check;
while (strcmp (search,tafel) != 0 ) {
i++;
letter = rand() % 26 +'A';
pos = rand() % len;
//pos=(pos+1)%len;
tafel[pos] = letter;
if (i%10000000==0) {
printf("%4ld mio: %s\n", i / 1000000, tafel);
printf ("%d\n",strcmp (search,tafel));
}
}
printf("It takes %ld trys to find '%s'\n",i,tafel);
printf ("%d\n",strcmp (suchwort,tafel));
return 0;
}
I need to get the length of the palindrome of the word in a string. Ex. tomyot length =2.
I wrote the following code but it doesn't work.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char str[20] = "tomyot";
char rstr[20];
strcpy(rstr, str);
strrev(rstr);
int i,j;
int count = 0;
int s=0;
for(i=0;i<strlen(str); i++){
for(j=s;j<strlen(str); j++){
if(str[i] == rstr[j]){
count+=1;
s = j+1;
continue;
}
}
}
printf("%d",count);
return 0;
}
Replace
sizeof(str)
with
strlen(str)
The former one returns the size of the str array which is 20 and the latter one returns the length of the contents of str which is 6.
I have made the changes and put comments in the code in /* .. */ blocks:
#include <stdio.h>
#include <string.h>
int main(void) {
/*
- You don't need to compute the reverse of the string, the input string itself will do your work.
- strrev() is not supported in GCC, so don't use it. See https://stackoverflow.com/a/8534275/4688321
for alternative implementation
*/
char str[20] = "tomyot";
int len_str = strlen(str);
int i, j, cnt = 0;
/*
- You don't need two nested for loops, one loop with two pointers:
first from the start of string and other from end of the string will do
- Just compare the ith character (from start) with the jth character (from end)
- Stop wherever i and j cross each other i.e. when i > j
*/
for (i = 0, j = len_str - 1; i <= j && i < len_str - 1; i++, j--) {
if (str[i] == str[j]) {
cnt++;
}
else break;
}
printf("%d\n", cnt);
return 0;
}
I am self teaching C programming.
I am trying to count number of int present in given string which are separated by space.
exp:
input str = "1 2 11 84384 0 212"
output should be: 1, 2, 11, 84384, 0, 212
total int = 6
When I try. It gives me all the digits as output which make sense since I am not using a right approach here.
I know in python I can use str.split (" ") function which can do my job very quickly.
But I want to try something similar in C. Trying to create my own split method.
#include <stdio.h>
#include <string.h>
void count_get_ints(const char *data) {
int buf[10000];
int cnt = 0, j=0;
for (int i=0; i<strlen(data); i++) {
if (isspace(data[i] == false)
buf[j] = data[i]-'0';
j++;
}
printf("%d", j);
}
// when I check the buffer it includes all the digits of the numbers.
// i.e for my example.
// buf = {1,2,1,1,8,4,3,8,4,0,2,1,2}
// I want buf to be following
// buf = {1,2,11,84384,0,212}
I know this is not a right approach to solve this problem. One way to keep track of prev and dynamically create a memory using number of non space digits encountered.
But I am not sure if that approach helps.
You want to build your number incrementally until you hit a space, then put that into the array. You can do this by multiplying by 10 then adding the next digit each time.
void count_get_ints(const char *data) {
int buf[10000];
int j = 0;
int current_number = 0;
// Move this outside the loop to eliminate recalculating the length each time
int total_length = strlen(data);
for (int i=0; i <= total_length; i++) {
// Go up to 1 character past the length so you
// capture the last number as well
if (i == total_length || isspace(data[i])) {
// Save the number, and reset it
buf[j++] = current_number;
current_number = 0;
}
else {
current_number *= 10;
current_number += data[i] - '0';
}
}
}
I think strtok will provide a cleaner solution, unless you really want to iterate over every char in the string. It has been a while since I did C, so please excuse any errors in the code below, hopefully it will give you the right idea.
#include <stdio.h>
#include <stdlib.h>
int main() {
char str[19] = "1 2 11 84384 0 212";
const char s[2] = " ";
char *token;
int total;
total = 0;
token = strtok(str, s);
while (token != NULL) {
printf("%s\n", token);
total += atoi(token);
token = strtok(NULL, s);
}
printf("%d\n", total);
return 0;
}
You can check the ascii value of each character by doing c-'0'. If it's between [0,9], then it's an integer. By having a state variable, when you're inside an integer by checking if a given character is a number of space, you can keep track of the count by ignoring white space. Plus you don't need a buffer, what happens if data is larger than 10,000, and you write pass the end of the buffer?, undefined behavior will happen. This solution doesn't require a buffer.
Edit, the solution now prints the integers that are in the string
void count_get_ints(const char *data) {
int count = 0;
int state = 0;
int start = 0;
int end = 0;
for(int i = 0; i<strlen(data); i++){
int ascii = data[i]-'0';
if(ascii >= 0 && ascii <= 9){
if(state == 0){
start = i;
}
state = 1;
}else{
//Detected a whitespace
if(state == 1){
count++;
state = 0;
end = i;
//Print the integer from the start to end spot in data
for(int j = start; j<end; j++){
printf("%c",data[j]);
}
printf(" ");
}
}
}
//Check end
if(state == 1){
count++;
for(int j = start; j<strlen(data); j++){
printf("%c",data[j]);
}
printf(" ");
}
printf("Number of integers %d\n",count);
}
I believe the standard way of doing this would be using sscanf using the %n format specifier to keep track of how much of the string is read.
You can start with a large array to read into -
int array[100];
Then you can keep reading integers from the string till you can't read anymore or you are done reading 100.
int total = 0;
int cont = 0;
int ret = 1;
while(ret == 1 && total < 100) {
ret = sscanf(input, "%d%n", &array[total++], &cont);
input += cont;
}
total--;
printf("Total read = %d\n", total);
and array contains all the numbers read.
Here is the DEMO
Example using strtol
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <errno.h>
#include <ctype.h>
int count_get_ints(int output[], int output_size, const char *input) {
const char *p = input;
int cnt;
for(cnt = 0; cnt < output_size && *p; ++cnt){
char *endp;
long n;
errno = 0;
n = strtol(p, &endp, 10);
if(errno == 0 && (isspace((unsigned char)*endp) || !*endp) && INT_MIN <= n && n <= INT_MAX){
output[cnt] = n;
while(isspace((unsigned char)*endp))
++endp;//skip spaces
p = endp;//next parse point
} else {
fprintf(stderr, "invalid input '%s' in %s\n", p, __func__);
break;
}
}
return cnt;
}
int main(void) {
const char *input = "1 2 11 84384 0 212";
int data[10000];
int n = sizeof(data)/sizeof(*data);//number of elements of data
n = count_get_ints(data, n, input);
for(int i = 0; i < n; ++i){
if(i)
printf(", ");
printf("%d", data[i]);
}
puts("");
}
Assuming you don't have any non-numbers in your string, you can just count the number of spaces + 1 to find the number of integers in the string like so in this pseudo code:
for(i = 0; i < length of string; i++) {
if (string x[i] == " ") {
Add y to the list of strings
string y = "";
counter++;
}
string y += string x[i]
}
numberOfIntegers = counter + 1;
Also, this reads the data between the white spaces. Keep in mind this is pseudo code, so the syntax is different.
i'm developing a little function to display the most frequent character in a (char) array.
This is what I've accomplished so far, but I think i'm on the wrong way.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char test[10] = "ciaociaoci";
max_caratt(test, 10);
}
int max_caratt(char input[], int size)
{
int i;
char max[300];
max[0] = input[0];
for (i=0; i<size; i++)
{
if(strncmp(input,input[i],1) == 1)
{
printf("occourrence found");
max[i] = input[i];
}
}
}
Any help?
Actually, the correct code is this.
It's just a corrected version of IntermediateHacker's below snippet.
void main()
{
int array[255] = {0}; // initialize all elements to 0
char str[] = "thequickbrownfoxjumpedoverthelazydog";
int i, max, index;
for(i = 0; str[i] != 0; i++)
{
++array[str[i]];
}
// Find the letter that was used the most
max = array[0];
index = 0;
for(i = 0; str[i] != 0; i++)
{
if( array[str[i]] > max)
{
max = array[str[i]];
index = i;
}
}
printf("The max character is: %c \n", str[index]);
}
The easiest way to find the most common character is to create an int array of 255 and just increment the arraly element that corresponds to the character. For example: if the charcter is 'A', then increment the 'A'th element (if you look at any ascii table you will see that the letter 'A' has a decimal value of 65)
int array[255] = {0}; // initialize all elements to 0
char str[] = "The quick brown fox jumped over the lazy dog.";
int i, max, index;
// Now count all the letters in the sentence
for(i = 0; str[i] != 0; i++)
{
++array[str[i]];
}
// Find the letter that was used the most
max = array[0];
index = 0;
for(i = 0; str[i] != 0; i++)
{
if( array[i] > max)
{
max = array[i];
index = i;
}
}
printf("The max character is: %c \n", (char)index);
You're passing a (almost) string and a char to strncmp(). strncmp() takes two strings (and an integer). Your program shouldn't even compile!
Suggestion: increase the warning level of your compiler and mind the warnings.
You may want to look at strchr() ...
Assuming an input array of 0-127, the following should get you the most common character in a single pass through the string. Note, if you want to worry about negative numbers, shift everything up by +127 as needed...
char mostCommonChar(char *str) {
/* we are making the assumption that the string passed in has values
* between 0 and 127.
*/
int cnt[128], max = 0;
char *idx = str;
/* clear counts */
memset((void *)cnt, 0, sizeof(int) * 128);
/* collect info */
while(*idx) {
cnt[*idx]++;
if(cnt[*idx] > cnt[max]) {
max = *idx;
}
idx++;
}
/* we know the max */
return max;
}
If you don't need to preserve the input array, you could sort the input array first, then find the longest contiguous run of a single character. This approach is slower, but uses less space.
I made a working version using structs. It works fine, I guess, but I think there's a MUCH better way to write this algorithm.
#include <stdio.h>
#include <stdlib.h>
struct alphabet {
char letter;
int times;
};
typedef struct alphabet Alphabet;
void main() {
char string[300];
gets(string);
Alphabet Alph[300];
int i=0, j=0;
while (i<=strlen(string)) {
while(j<=300) {
if(string[i] != Alph[j].letter) {
Alph[i].letter = string[i];
Alph[i].times = 1;
}
else {
Alph[j].times++;
}
j++;
}
j=0;
i++;
}
int y,max=0;
char letter_max[0];
for (y=0; y<strlen(string); y++) {
printf("Letter: %c, Times: %d \n", Alph[y].letter, Alph[y].times);
if(Alph[y].times>max) {
max=Alph[y].times;
letter_max[0]=Alph[y].letter;
}
}
printf("\n\n\t\tMost frequent letter: %c - %d times \n\n", letter_max[0], max);
}
I saw you all creating big arrays and "complex" stuff so here I have easy and simple code xD
char most_used_char (char s[]) {
int i; //array's index
int v; //auxiliary index for counting characters
char c_aux; //auxiliary character
int sum = 0; //auxiliary character's occurrence
char c_max; //most used character
int max = 0; //most used character's occurrence
for (i = 0; s[i]; i++) {
c_aux = s[i];
for (v = 0; s[v]; v++)
if (c_aux == s[v]) sum++; /* responsible cycle for counting
character occurrence */
if (sum > max) { //checks if new character is the most used
max = sum;
c_max = c_aux;
}
sum = 0; /* reset counting variable so it can counts new
characters occurrence */
}
return c_max; //this is the most used character!
}