I have a homework problem. It requires me to convert a word into uppercase and several characters have to be converted to decimal for example :
"Hello my NamE is FeLix" --> "H3LL0 MY N4M3 15 F3L1X". So, these characters had to be converted :
I = 1
S = 5
E = 3
O = 0
A = 4
etc.
How to convert it? I already tried to convert it to capslock but i cannot convert it into decimal.
I already tried to convert the words into uppercase, but have no idea how to convert the character into numbers.
int main()
{
char sentence[200];
int sentencelength = strlen(sentence);
// Ambil data user
scanf("%s",&sentence); getchar();
// Cek satu persatu pake for
for (int i= 1; i <= sentencelength; i++) {
if(sentence[i] >= 'a' && sentence[i] <= 'z') {
char uppercase = sentence[i] + 'A' - 'a';
printf("%c",uppercase);
}
}
getchar();
return 0;
}
There is no error, but I just have no idea how to convert it.
You can use switch as below.
switch(uppercase ) {
case 'I':
uppercase = '1';
break;
case 'S':
uppercase = '5';
break;
case 'E':
uppercase = '3';
break;
…
}
C arrays start at index 0, not 1, so change the for loop bounds to:
for (int i = 0; i < sentencelength; i++) {
You can use toupper (declared by #include <ctype.h>) to convert a character from lowercase to uppercase, leaving non-alphabetic characters alone. It is only defined for values representable by an unsigned char or for the value EOF.
char l33t = sentence[i];
if (l33t == (unsigned char)l33t)
l33t = toupper(l33t);
You can use a switch statement to replace certain uppercase letters with digits:
switch (l33t) {
case 'I':
l33t = '1';
break;
case 'S':
l33t = '5';
break;
case 'E':
l33t = '3';
break;
case 'O':
l33t = '0';
break;
case 'A':
l33t = '4';
break;
}
Rather than using scanf to read a whole word of input into a buffer, an alternative is to read the input a character at a time. Here is an example program that behaves as a filter:
#include <stdio.h>
#include <ctype.h>
static int convert(int ch)
{
if (ch == (unsigned char)ch)
ch = toupper(ch);
switch (ch) {
case 'I': ch = '1'; break;
case 'S': ch = '5'; break;
case 'E': ch = '3'; break;
case 'O': ch = '0'; break;
case 'A': ch = '4'; break;
}
return ch;
}
int main(void)
{
int ch;
while ((ch = getchar()) != EOF) {
ch = convert(ch);
putchar(ch);
}
return 0;
}
The above will convert the whole input until it sees end-of-file. To terminate after a single line, just add a check for a newline character to break out of the while loop.
create an array of characters: [4BCD3F....Z]
and an array of sources: [abcd...z]
run on your string, replace each character found in index I with the same character in the first array, if it's not found return the character as is.
crude, simple, works
Also, if someone complain on the calculation complexity, since you have fixed number of letters in the arrays A to Z , the complexity is O(N*M) when M is const, hence O(N) anyway
Related
I need to replace some chars with multiple chars (a string), but I got stuck. This code works for replacing one char with another, but if the replacement are multiple chars, the output messes up.
Here is the code I have so far:
char input[50];
char output[150];
int i;
printf("Enter your text: ");
fgets(input, 50 , stdin);
for (i = 0; input[i] != '\0'; i++){
switch (input[i]){
case 'a': output[i]= '4'; break;
case 'd': output[i]= '|)';break;
case 'e': output[i]= '3'; break;
case 'f': output[i]= '|='; break;
case 'u': output[i]= '|_|'; break;
case 'w': output[i]= '\|/'; break;
/* REST OF THE ALPHABET
INTENTIONALLY SUPPRESSED*/
}
}
printf("Your new text is: %s", output);
return 0;
As suggested by dasblinkenlight, I set another index for the output, That worked pretty fine, but I getting two additional chars at the end of the output text... where does those chars come from?
This is an example:
Enter your text: afedef
Your new text is: 4|=3|)3|=■(
1) You can't copy strings with = operator.
2) String literal in c is surrounded with double quotes
3) Special characters like backslash ('\') in string literals have a special meaning and should be escaped.
4) In order to do what you intend you will need:
a) Additional counter to track the position in the output string, where the next string will be written
b) use of strcpy/strncpy instead of assignment operator = , such that each line in the `switch statement will look similar to this:
case 'f': strcpy(&output[j], "|="); j+=2; break;
Here j is the second counter incremented by the number of characters written to output
Take a look at strcat (http://www.cplusplus.com/reference/cstring/strcat/):
char input[50];
char output[150];
int i;
printf("Enter your text: ");
fgets(input, 50 , stdin);
output[0] = 0; // null terminate before calling strcat.
for (i = 0; input[i] != '\0'; i++)
{
switch (input[i])
{
case 'a': strcat(output, "4"); break;
case 'd': strcat(output, "|)"); break;
//...
}
}
printf("Your new text is: %s", output);
Also, multiple characters in single quotes in C++ is implementation defined (C++ multicharacter literal), you'll probably want to avoid them. Also, be careful when using '\' in literals (as with your last case), it's probably not doing what you expect.
There are two problems in your code:
you are using i to iterate the output array. The size of the input array and output array are different. It may happen (e.g. in the case of f) that the index for the input array needs to be incremented by 1, whereas the index for the output array should be incremented by 2 or 3.
you can't assign multiple chars to a single char slot in an array. For instance, case 'f': output[i]= '|='; break; is incorrect.
To solve the problem you should use another variable and increment it with the number of characters added. For example:
int j = 0;
...
case 'e':
output[j++]= '3';
break;
case 'f':
output[j++]= '|';
output[j++]= '=';
break;
Try the following
#include <string.h>
//...
char *p = input, *q = output;
do
{
switch ( *p )
{
case '\0': *q = *p; break;
case 'a': *q++ = '4'; break;
case 'd': strcpy( q, "|)" ); q += 2; break;
case 'e': *q++ = '3'; break;
case 'f': strcpy( q, "|=" ); q += 2; break;
case 'u': strcpy( q, "|_|" ); q += 3; break;
case 'w': strcpy( q, "\|/" ); q += 3; break;
/* REST OF THE ALPHABET
INTENTIONALLY SUPPRESSED*/
}
} while ( *p++ );
Or instead of using strcpy you can assign each caharcter one after another as for example
case 'd': *q++ = '|'; *q++ = ')'; break;
I'm new to C and I've been working on this homework problem for about 2 hours to no avail. I'm attempting to create a program that takes an alphabetic phone number (ie; CALLATT or 1-800-COL-LECT) and turns it into the number form (2255288 or 1-800-265-5328). No matter what I put for input, though, I always get -4197680 for my output.
int main(void){
int c=0, len, a[len];
char n[len];
printf("Enter phone number: \n");
scanf("%c", n);
len = sizeof(n) / sizeof(n[0]);
while (len > c){
if (n[c] == 'A' || n[c] == 'B' || n[c] == 'C'){
a[c] = 2;
c++;
}
else if (n[c] == 'D' || n[c] == 'E' || n[c] == 'F'){
a[c] = 3;
c++;
}
else if (n[c] == 'G' || n[c] == 'H' || n[c] == 'I'){
a[c] = 4;
c++;
}
else if (n[c] == 'J' || n[c] == 'L' || n[c] == 'L'){
a[c] = 5;
c++;
}
else if (n[c] == 'M' || n[c] == 'N' || n[c] == 'O'){
a[c] = 6;
c++;
}
else if (n[c] == 'P' || n[c] == 'Q' || n[c] == 'R' || n[c] == 'S'){
a[c] = 7;
c++;
}
else if (n[c] == 'T' || n[c] == 'U' || n[c] == 'V'){
a[c] = 8;
c++;
}
else if (n[c] == 'W' || n[c] == 'X' || n[c] == 'Y' || n[c] == 'Z'){
a[c] = 9;
c++;
}
else {
a[c] = n[c];
c++;
}
}
printf("%d\n", a);
return 0;
}
EDIT: Revised. There were many comments pointing out problems, here is my answer which works with a reasonable length phone number. It skips any non-dialing characters, such as '-' which is not part of a phone number.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void){
int k, d, e, len;
char dial[20], entry[20] = {0};
printf("Enter phone number: ");
fgets(entry, 19, stdin);
len = strlen(entry);
d = 0; // dial string index of output
for (e=0; e<len; e++) { // entry string index of input
k = entry[e];
switch (toupper(k)) {
case 'A': case 'B': case 'C': dial[d++] = '2'; break;
case 'D': case 'E': case 'F': dial[d++] = '3'; break;
case 'G': case 'H': case 'I': dial[d++] = '4'; break;
case 'J': case 'K': case 'L': dial[d++] = '5'; break;
case 'M': case 'N': case 'O': dial[d++] = '6'; break;
case 'P': case 'Q': case 'R': case 'S': dial[d++] = '7'; break;
case 'T': case 'U': case 'V': dial[d++] = '8'; break;
case 'W': case 'X': case 'Y': case 'Z': dial[d++] = '9'; break;
default:
if (isdigit(k) || k=='*' || k=='#') dial[d++] = k;
}
}
dial[d] = 0; // terminate string
printf("Dial %s\n", dial);
return 0;
}
Here is some code:
char buf[32];
sscanf("%31s", buf);
size_t i;
for (i = 0; i < sizeof(buf) && buf[i]; ++i)
{
switch (buf[i])
{
case 'A': case 'B': case 'C':
buf[i] = '2'; break; // Note: character literal, not integer
case 'D': case 'E': case 'F':
buf[i] = '3'; break;
....
}
}
printf("%s", buf);
If you have a Posix-compliant library, you can use dynamic allocation:
char *buf;
scanf("%ms", &buf); //scanf would allocate memory
for (i = 0; buf[i]; ++i)
{
.....
}
printf("%s", buf);
free(buf);
There are so many problems in your code, it will almost need a re-write to make it work. I think you should start something small. Make sure it works before adding more functionality. I would suggest dividing the code in main into three sections -- reading the phone number, converting phone number and printing the converted phone number.
Here's a skeletal program that captures those three steps.
#define SIZE 50
void readPhoneNumber(char phoneNumber[])
{
}
void convertTextToNumber(char phoneNumber[], char dialedNumber[])
{
}
void printPhoneNumber(char phoneNumber[])
{
}
int main(void)
{
char phoneNumber[SIZE];
char dialedNumber[SIZE];
readPhoneNumber(phoneNumber);
convertTextToNumber(phoneNumber, dialedNumber);
printPhoneNumber(dialedNumber);
}
Now, you can start fleshing out the functions. For example, readPhoneNumber can be implemented as:
void readPhoneNumber(char phoneNumber[])
{
printf("Enter phone number: \n");
fgets(phoneNumber, SIZE, stdin);
}
printPhoneNumber can be implemented as:
void printPhoneNumber(char phoneNumber[])
{
printf("%s\n", phoneNumber);
}
I'll leave you to work out the implementation of convertTextToNumber.
Here you have undefined behavior, len is not initialized.
int c=0, len, a[len];
char n[len];
Use instead a constant value instead, i bet the phone number in your country has some kind of maximum length.
This way to read from the keyboard is not recommended, scanf does not check for length of string so you can do a faceroll on the keyboard and your program will crash. Instead use fgets( ) to read from stdin then go through the string char by char skipping the included \n
printf("Enter phone number: \n");
scanf("%c", n);
This makes no sense, you calculate the sizeof n i.e. of the integer that holds n. If you want the length of the string use strlen( n ); btw try to use more descriptive variable names.
len = sizeof(n) / sizeof(n[0]);
Instead of
while (len > c){
why not use a normal for-loop ? you seem to increment c++ everywhere.
this here will not do what you expect it to do
printf("%d\n", a);
but you assign 'a' integers e.g.
a[c] = 2;
printf can not magically print a number of your array, instead you want to print out is a string with the numbers. the ascii value of a digit is 48 + digit. e.g. '0' is 48, by knowing this have a character buffer and add the ascii values to it. make sure it ends with \0 which is end of string. then print out the string
buf[c++] = 48 + digit;
...
buf[c] = '\0';
puts( buf );
char convertalphas(char s) {
switch (s){
case 'A':
return '0';
break;
case 'B':
return '1';
break;
case 'C':
return '2';
break;
case 'D':
return '3';
break;
case 'E':
return '4';
break;
case 'F':
return '5';
break;
case 'G':
return '6';
break;
case 'H':
return '7';
break;
case 'I':
return '8';
break;
case 'J':
return '9';
break;
case 'K':
return '10';
break;
case 'L':
return '11';
break;
case 'M':
return '12';
break;
case 'N':
return '13';
break;
case 'O':
return '14';
break;
case 'P':
return '15';
break;
case 'Q':
return '16';
break;
case 'R':
return '17';
break;
case 'S':
return '18';
break;
case 'T':
return '19';
break;
case 'U':
return '20';
break;
case 'V':
return '21';
break;
case 'W':
return '22';
break;
case 'X':
return '23';
break;
case 'Y':
return '24';
break;
case 'Z':
return '25';
break;
}
}
int main()
{
char astring[10];
int i = 0;
int flag = 0;
int startedalpha = 0;
//check if there is a digit input or not
int nodigit = 0;
char cell[10];
int col;
scanf( "%s", &astring );
for ( i = 0; i < 10; ++i )
{
if(astring[i] != '\0') {
//check whether letter is capital or small
if (astring[i] >= 65 && astring[i] <= 90)
{
startedalpha = 1;
//printf( "%c\n", astring[i] );
cell[i] = convertalphas(astring[i]);
printf("cell is %s\n", cell);
}
What im trying to do is to concatenate all of my conversions for a later use. when I put "AB" it returns "01#" and when I put "Z" it returns something else than "25". I don't know what is wrong but it is driving me crazy! I want to be able to input "ABZ" and it saves all of my values into the variable cell. For example, "ABZ" "0125"
Thanks!
I'm confused with this forum. Obviously, I posted the question because there is something I don't know and I'm not a guru as many of you! So, why would I get -2? I already posted what I tried. I thought it is about helping not being condescending!
Thanks for those who replied anyway!
Edit --
I converted my switch statement to int but now how can I concatenate the integers in variable cell?
One of things I noticed is that you have a char returning function, and is returning something else than a char, like '10'. It will return an unexpected value. You can, for example, set the return of this function to integer.
So it will look like this:
int convertalphas(char s) {
switch (s){
case 'A':
return 0;
case 'B':
return 1;
case 'C':
return 2;
case 'D':
return 3;
case 'E':
return 4;
return -1;
}
And then, change inside the loop in main function:
sprintf(auxvar, "%d", convertalphas(astring[i]);
for( jj = 0; jj < strlen(auxvar); jj++)
cell[i++] = auxvar[jj];
I hope it can help you somehow!
Good Luck.
According to the C Standard (6.4.4.4 Character constants)
...The value of an integer character constant containing more than one
character (e.g., 'ab'), or containing a character or escape sequence
that does not map to a single-byte execution character, is
implementation-defined.All
All return values of the function starting from '10' to '25' inclusively are imolementation defined and you can get the result that you did not expect to get.
You need to write a function that would be declared like
char * convertalphas( const char *s );
Or
char * convertalphas( char *dest, const char *source );
By the way it will be difficult to make the reverse conversion. For example what does "25" mean? Whether it is "CF" or "Z"?:)
Apart from the values returned by chars 'K' thru 'Z' (as mentioned by others) there are some other errors with the use of cell
Each character returned from your function is written into cell[] and then you print cell as a string. But there is no string terminator. You either need to fill cell[] with zeros first, or write a 0 into the next character
cell[i] = convertalphas(astring[i]);
cell[i+1] = 0;
But the problem with this is that cell[i] is skipped when the char is not a capital letter. You need to maintain a separate index into cell[]
cell[j++] = convertalphas(astring[i]);
cell[j] = 0;
One more problem is cell[] is not long enough for the final string terminator. You need to declare it as
char cell[11];
To summarize, your problem is:
A char function cannot return anything other than a single character. If you try to return something like '12', it just simply doesn't work. You need to rethink your algorithm. Implementing this using ints should be very straightforward.
Is this a homework assignment requiring the use of the switch statement? If not, and if you follow the advice to return int instead of char, a simple s-65 would do the trick, you don't need a 78-line switch.
Since cell is a char array, each element of cell can also only store a single character.
It is not very clear what you are trying to achieve, so I will try to comprehend it in two ways:
3.1. You want to store each character's value separately in cell.
Example: for input "ABZ", you want cell to be {0, 1, 25}.
This is easy. Change cell to an int array, and output them using %d format specifier in your printf.
3.2 You treat cell as a string rather than an array of characters, and you concatenate the string for every character's value.
Example: for input "ABZ", you want cell to be "0125", or in other words, {'0', '1', '2', '5'}. This means that you won't be able to differentiate between "ABZ" and, say, "AMF", as others have pointed out.
If this is what you want to achieve, firstly 10 elements is not enough for cell - each character must be stored separately, and you cannot store "25" as two characters in a single element of cell.
You can assign it with, say, 21 elements (2 max for each alphabet, one extra for the terminating '\0' byte). After each alphabet is converted to its value in int, implement a counter and some conditionals to fill cell one character at a time. Example, if an alphabet's value is stored in val:
char cell[21];
int i=0;
if(val < 10) {
cell[i] = val + '0';
i++;
} else {
cell[i] = val / 10 + '0';
i++;
cell[i] = val % 10 + '0';
i++;
}
I'll leave the implementation of this into a loop as an exercise for you...
I'm new in C and I couldnt find the answer to my question in the forum.
The point is, I need to get a value of deck cards from the user. So it can spread from 2 to 10 and also be 'J', 'Q', 'K' or 'A'. That means it can be a integer or a character.
I'm trying to put it in an integer variable called "_val1". This work for any number from 0 to 10. I expected that if I typed a letter, _val1 would get the ASCII value of that character (wich I could use later for my pourposes). But instead _val1 geta value '0' and the letter is automatically passed to my next variable call (wich is _naipe1).
How can I solve that?
That means, how cam I use scanf to get either a integer value or the ASCII value of a character?
short int _val1, _val2;
char _naipe1, _naipe2;
printf("Qual a 1ª carta?\n Valor:");
scanf(" %hd", &_val1);
printf("Valor 1 = %hd \n", _val1 );
printf(" Naipe:");
scanf(" %c", &_naipe1);
well, if I were you I'd try to simplify the problem:
get the ASCII value of the card representation from '2' to '9' and 'J','Q','K','A' ; there you can simply use a scanf("%c") or even better a getchar() operation.
then either you keep using the ASCII representation of your cards throughout your algorithm, or you can translate it using a mapping function such as:
int map(char card) {
switch (card) {
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
return card-'0';
case 'A':
return 1;
case 'J':
return 10;
case 'Q':
return 11;
case 'K':
return 12;
}
}
First, there are 52 cards to a typical poker deck, These are split into 4 suits: hearts, diamonds, spades and clubs. This kind of suggests that user input will be something like: 10s, Ad, 3c, etc (meaning 10 of spades, Ace of diamonds and 3 of clubs) So, not only must you determine the value of the individual card, you also must determine the suit.
This will not solve all of those requirements, but it will at least answer your most direct question, how to read an int or a char using scanf().
This will demonstrate that:
#include <stdio.h>
int main(int argc, char** argv)
{
int aNumber;
char aChar;
printf("\nEnter a number:");
scanf("%d", &aNumber);
printf("\nEnter a character:");
scanf("%c", &aChar);
printf("\nThe number entered is %d\n", aNumber);
printf("\nThe character entered is %c\n", aChar);
return 0;
}
You can also simply have all the values in a string such as
char cards[]={"Ad Js 10c 2c Qh"};
Then parse it using strtok(), then test each token for its ascii content, using functions like isdigit() or isalpha()
Note: you will have to map each card to a value to keep them straight, something like this abbreviated enum may work:
enum {
AD = 1, //start enum values at 1 for the diamonds suit
2D,
3D,
...//fill in rest of cards here
JC,
QC,
KC, // last card == 52, with the clubs suit
};
The reason your output from _val1 is 0 when entering a letter lies in the fact that you've declared _val1 as an short int. You should be using a char. Then you can assign and compare their ascii values.
char card;
int value;
scanf("%c", card);
if(card < 58 && card > 49)
value = card - 48;
else {
switch(card) {
case 'a': value = 1;
case '0': value = 10;
case 'j': value = 11;
case 'q': value = 12;
case 'k': value = 13;
default: printf("Must enter 0-9 (0 for 10 card), or a, j, q, k\n");
}
}
To read in "A", "2", "3", ... "10", "J",... "K", use fgetc() and strchr().
#include <ctype.h>
#include <stddef.h>
#include <stdio.h>
#include <string.h>
short GetCardRank(void) {
static const char rank[] = "A234567891JQK";
short val = -1;
int ch = fgetc(stdin);
while (isspace(ch)) ch = fgetc(stdin); // Skip leading white-space
char *p = strchr(rank, toupper(ch)); // Use toupper() to make case insensitive
if (ch != EOF && p != NULL && *p != '\0') {
short val = (short) (p - rank + 1);
if (val != 10) return val;
ch = fgetc(stdin);
if (ch == '0') return val;
val = 1; // Allow a lone '1' to act like an 'A'
}
ungetc(ch, stdin); // Put back unused char for next IO function
return val;
}
I'm trying to put it in an integer variable called "_val1". This work for any number from 0 to 10. I expected that if I typed a letter, _val1 would get the ASCII value of that character (wich I could use later for my pourposes). But instead _val1 geta value '0' and the letter is automatically passed to my next variable call (wich is _naipe1)
The problem is that the %d conversion specifier only recognizes strings of decimal digits (with an optional leading + or -) and will stop reading at the first non-digit character; if you type in something other than a digit, then the input operation will fail and that character will be left in the input stream.
Your best bet is to read your input as text, then convert it to a numerical value manually, something like the following:
#include <ctype.h>
#include <stdlib.h>
/**
* Reads a card's face value (2-10,J,Q,K,A) from standard input
* Returns 0 on error
*/
short get_card_value( void )
{
char buf[4]; // large enough to hold a 2-digit string plus newline plus 0 terminator
short val = 0;
if ( fgets( buf, sizeof buf, stdin ) != NULL )
{
char *chk;
short tmp = (short) strtol( buf, &chk, 0 );
if ( isspace( *chk ) || *chk == 0 )
{
if ( tmp >= 2 && tmp <= 10 )
val = tmp;
}
else
{
switch( tolower( *chk ) )
{
case 'j': val = 11; break;
case 'q': val = 12; break;
case 'k': val = 13; break;
case 'a': val = 11; break;
default: break;
}
}
}
// else read error
return val;
}
You'd call this as
val1 = get_card_value();
if ( val1 == 0 )
// error on input
This code doesn't do any length checking on input, so if you enter a card value of 1234567890, that won't be handled gracefully.
Don't use leading underscores in your variable names; names with leading underscores are reserved for the implementation.
I am new a C. I would like to get help to finish my function.
The mission is:
Write a function that accepts a string maximum length of 256 characters containing characters from 'a' to 'z'.
The function to print the number of occurrences of each character.
For example: input abba output will be:
a = 2 b = 2 c = 0 d = 0 .... z = 0
Do not use if during any function.
I would like to get your help to finish this program please.
This is my code
#include "stdlib.h"
#include "conio.h"
#include "stdio.h"
#include "string.h"
#define size 256
void repeat(char *str);
void main()
{
char str[size];
printf("Please enter a string:\n");
flushall;
gets(str);
repeat(str);
system("pause");
return ;
}
void repeat(char *str)
{
char temp=strlen(str);
int i, count=0;
do
{
for (i=0; i<temp ; i++)
{
count += (*str == str[temp-i]);
}
printf("Char %c appears %d times\n ",*str,count);
count=0;
}
while(*(str++));
}
Please enter a string:
abbba
Char a appears 1 times
Char b appears 2 times
Char b appears 1 times
Char b appears 0 times
Char a appears 0 times
Char appears 0 times
Press any key to continue . . .
this is the output!
I would like to do it in the same building i did.
and should be like
Char a appears 2 times
Chars b appears 3 times
You make a stipulation about not using if. This satisfies that restriction.
#include <stdio.h>
int main(void) {
int i, c;
int counts[256] = { 0 };
const char lower[] = "abcdefghijklmnopqrstuvwxyz";
while ((c = getchar()) != EOF) {
counts[c] += 1;
}
for (i = 0; lower[i]; ++i) {
c = lower[i];
printf("Char %c appears %d times.\n", c, counts[c]);
}
return 0;
}
The problem with your attempt is that you do not track any state to remember which characters you have already printed information about. It also fails to include the character under consideration as part of the count. It also makes multiple passes over the string to collect count information about each character, but that doesn't affect correctness, just performance. If you can somehow remember which character you have already printed out information for, so that you don't do it again when the same character appears later in the string, your method should print out the counts for the characters that appear. Afterwards, you would need to print out zero counts for the characters that did not appear at all. If the outputs need to be in alphabetical order, then you need to make sure you take care of that as well.
One way to track the information properly and to allow your output to be printed in alphabetical order is to maintain counts for each character in an array. After making a pass over the string and incrementing the count associated with each found character, you can iterate over the count array, and print out the counts.
The following program is for zubergu:
#include <stdio.h>
#include <string.h>
int main (void) {
int i, c;
int counts[26] = { 0 };
const char lower[] = "abcdefghijklmnopqrstuvwxyz";
while ((c = getchar()) != EOF) {
switch (c) {
case 'a': case 'b': case 'c': case 'd': case 'e': case 'f': case 'g':
case 'h': case 'i': case 'j': case 'k': case 'l': case 'm': case 'n':
case 'o': case 'p': case 'q': case 'r': case 's': case 't': case 'u':
case 'v': case 'w': case 'x': case 'y': case 'z':
counts[strchr(lower, c) - lower] += 1;
break;
default:
break;
}
}
for (i = 0; lower[i]; ++i) {
printf("Char %c appears %d times.\n", lower[i], counts[i]);
}
return 0;
}
It might be one of the ugliest solutions, but also the simplest:
while(*str!='\0')
{
switch(tolower(*str))
{
case 'a': a_count++;break;
case 'b': b_count++;break;
.
.
.
}
str++;
}
It checks if str points to valid letter, then turns it to lower, so it's not case sensitive('A' will be same as 'a' character). No 'if' used and will work with every length char array terminated with '\0' char.
EDIT I have edited the program to follow the requirements of #SagiBinder.
(In my old version, I used an if sentence that checked if the character is in the set 'a'...'z').
The type of temp must be "bigger", that is, something different to char.
Try int, instead.
The algorithm would be this (some details of your program are not repeated here):
int temp = strlen(str);
int i, j;
unsigned char c;
int ch[UCHAR_MAX]; // The macro CHAR_MAX needs the header <limits.h>
for (i = 1; i <= UCHAR_MAX; i++)
ch[i] = 0;
for (j=0; j<temp ; j++) {
c = (unsigned char)(str[j]);
ch[c]++;
}
for (c = 'a'; c <= 'z'; c++)
printf("%c == %d\n", c, ch[c]);
The variable temp holds the length of the string str.
The macro UCHAR_MAX (existing in the header <limits.h>, that you have to #include at the beginning of the program). It is the max. value that holds in a unsigned char.
The array ch[] contains a component for each possible value in the range of the type unsigned char. The intent is that, for some character c, the element ch[c] is the amount of times that c is in str.
I have used unsigned char in order to ensures that the index c of the array ch[] when writting ch[c] is a non-negative integer value, because an array cannot have negative indexes.
The 2nd for goes through the string str. In the step number j, the j-th character of the string str is taken.
This character is a value of type char.
Since one cannot be sure that char have not negative values, I have converted it to (unsigned char) with an explicit cast.
This value is held in the variable c.
The value of c has the (unsigned char version of the) j-th character in str,
so we are going to count it.
How?
Well, we access the array of counters: ch[] with index c, and increment its value in 1:
ch[c]++;
After the for is finished, we have in the array ch[] the information we want.
Finally, we check for the characters from 'a' to 'z'.
(For this, we have supposed that the character encodings in our system follow the convention that the letters have contiguous values).
The 3rd for goes from 'a' to 'z', and the values of the letter (the variable c that controls the for) and the counting of this letter, that is, ch[c].
Moreover: to show the count of any character, you need a re-cast to char, in this way:
printf("%c: %d\n", (char)c, ch[c]);
But this is not necessary with the letters 'a' to 'z', because they belong to the basic execution character set which means that their values are non-negative and equal to their unsigned char counterparts. So, in this case, it is enough to write:
printf("%c: %d\n", c, ch[c]);
EDIT 2: I will use the idea in the answer of #jxh to improve my code.
Since it cannot be guaranted that the encodings of letters 'a' to 'z' are in contiguous order, we can use a string holding the letters:
char letters[] = "abcdefghijklmnopqrstuvwxyz";
The "last" element is, by C convention, a \0 character held after the element 'z'.
Now, we can show the letter counting by changing the 3rd `for` in this way:
for (i = 0; letter[i] != '\0'; i++)
printf("%c == %d\n", letter[i], ch[letter[i]]);
This is equivalent to write:
for (i = 0; letter[i] != '\0'; i++) {
c = letter[i];
printf("%c == %d\n", c, ch[c]);
}
Optimized solution. complexity O(N), N - Input String length.
your void repeat function will be like this,
void repeat(char *str)
{
int temp=strlen(str);// use int here
int i, count=0;
int charCount[26] = {0};
#if 0
//your logic, traverses the string (n*n) time, n - input string length.
do
{
for (i=0; i<temp ; i++)
{
count += (*str == str[temp-i]);
}
printf("Char %c appears %d times\n ",*str,count);
count=0;
}
while(*(str++));
#endif
#if 1
// This logic traverses string once only. n time, n - input string length.
for (i=0; i<temp ; i++)
{
charCount[str[i]%'a']++;
}
for (i=0; i<26 ; i++)
{
printf("%c appears : %d times \n", 'a'+i, charCount[i]);
}
#endif
}
[EDIT]
Here
charCount[str[i]%'a']++; // 'a' is used a its ASCII Value.
You can use it as
charCount[str[i]%97]++;
If you wan to count lower case letter and upper case letter both.
use it like this
if(str[i] >= 'a' && str[i] <= 'z'){
iMap = str[i]%97; // 97 is ASCII Value of 'a'
charCount[iMap]++;
}else if(str[i] >= 'A' && str[i] <= 'Z'){
iMap = str[i]%65; // 65 is ASCII Value of 'A'
charCount[iMap]++;
}
//iMpa is a integer (int iMap;), used for better undersanding.
i = 0;
while (s[i] !=0)
if (( s[i] >= 'a' && s[i] <= 'z') || (s[i] <= 'A' && s[i] >= 'Z'))
{
letters++;
i++;
}
else
if (( s[i] >= '!' && s[i] <= ')'))
{
other++;
}
else
if (( s[i] >= '0' && s[i] <= '9'))
{
numbers++;
}
total = letters + numbers + other;