String from user input that gets converted into a letter pattern. - c

Question

Reposting because my first post was no good. I have a question that I'm not really sure how to do. I know the process I'm going for, but am not totally sure how to scan a string into an array so that each character/integer is scanned into a independent element of the array. I'll post the question and the code I have so far, and any help would be appreciated.
Question:
Assume that we have a pattern like the following: ([n][letter])+ in which n is an integer number and letter is one of the lowercase letters from a-z. For example, 2a and 3b are valid expressions based on our pattern. Also, “+” at the end of the pattern means that we have at least one expression (string) or more than one expression attached. For instance, 2a4b is another valid expression which is matched with the pattern. In this question, we want to convert these valid expressions to a string in which letters are repeated n times.
o Read an expression (string) from user and print the converted version of the expression in the output.
o Check if input expression is valid. For example, 2ab is not a valid expression. If the expression is not valid, print “Invalid” in the output and ask user to enteranother expression.
o Sample input1 = “2a”, output = aa
o Sample input2 = “2a3b”, output = aabbb
o You will receive extra credit if you briefly explain what concept or theory you can use to check whether an expression is valid or not.
What I have so far:
#include <stdio.h>
int main()
{
int size, i, j;
char pattern[20];
char vowel[20];
int count[20];
printf("Please enter your string: ");
gets(pattern);
size = strlen(pattern);
for(i=0; i<size; i++)
if((i+1)%2 == 0)
vowel[i] = pattern[i];
else if((i+1)%2 != 0)
count[i] = pattern[i];
for(i=0; i<size/2; i++);
for(j=0; j<count[i]; j++)
printf("%s", vowel[i]);
}

Answer 1

I assumed you want to write the "invalid\n" string on stderr. If not just change the file descriptor given to write.
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#define MAX_INPUT_SIZE 20
int
check_input(char *input)
{
while (*input)
{
if (*input < '0' || *input > '9')
{
write(2, "invalid\n", 8);
return 1;
}
while (*input >= '0' && *input <= '9')
input++;
if (*input < 'a' || *input > 'z')
{
write(2, "invalid\n", 8);
return 1;
}
input++;
}
return 0;
}
void
print_output(char *input)
{
int i;
while (*input)
{
i = atoi(input);
while (*input >= '0' && *input <= '9')
input++;
for (; i > 0; i--)
write(1, input, 1);
input++;
}
write(1, "\n", 1);
}
int
main()
{
char input[MAX_INPUT_SIZE];
do
{
printf("Please enter your string: ");
fgets(input, MAX_INPUT_SIZE, stdin);
input[strlen(input) - 1] = '\0';
}
while (check_input(input));
print_output(input);
return 0;
}

Answer 2

The steps are:
Read pattern
Check if pattern is valid
Generate output
Since the input length is not specified you have to assume a maximum length.
Another assumption is n is a single digit number.
Now you may read the whole expression with fgets() or read it char by char.
The latter allows you to check for validity as you read.
Lets use fgets() for convenience and in case the expression needs to be stored for later use.
char exp[100]; // assuming at most 50 instances of ([n][letter])
int len;
printf("Input: ");
fgets(exp, 100, stdin);
len = strlen(exp) - 1; // Discard newline at end
An empty input is invalid. Also a valid expression length should be even.
if (len == 0 || len%2 != 0) {
printf("Invalid-len\n");
return 1;
}
Now parse the expression and separately store numbers and letters in two arrays.
char nums[50], letters[50];
invalid = 0;
for (i = 0, j = 0; i < len; i += 2, j++) {
if (exp[i] >= '1' && exp[i] <= '9') {
nums[j] = exp[i] - '0';
} else {
invalid = 1;
break;
}
if (exp[i+1] >= 'a' && exp[i+1] <= 'z') {
letters[j] = exp[i+1];
} else {
invalid = 1;
break;
}
}
Notice that in each iteration if first char is not a number or second char is not a letter, then the expression is considered to be invalid.
If the expression is found to be invalid, nothing to do.
if (invalid) {
printf("Invalid\n");
return 1;
}
For a valid expression run nested loops to print the output.
The outer loop iterates for each ([n][letter]) pattern.
The inner loop prints n times the letter.
printf("Output: ");
for (i = 0; i < len/2; i++) {
for (j = 0; j < nums[i]; j++)
printf("%c", letters[i]);
}
This is a rather naive way to solve problems of this type. It is better to use regular expressions.
C standard library doesn't have regex support. However on Unix-like systems you can use POSIX regular expressions.

Answer 3

like this
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>
#define prompt "Please enter your string: "
void occurs_error(const char *src, const char *curr){
printf("\nInvalid\n");
printf("%s\n", src);
while(src++ != curr){
putchar(' ');
}
printf("^\n");
}
bool invalid(char *pattern){
char *p = pattern;
while(*p){
if(!isdigit((unsigned char)*p)){//no number
occurs_error(pattern, p);
break;
}
strtoul(p, &p, 10);
if(!*p || !islower((unsigned char)*p)){//no character or not lowercase
occurs_error(pattern, p);
break;
}
++p;
}
return *p;
}
int main(void){
char pattern[20];
while(fputs(prompt, stdout), fflush(stdout), fgets(pattern, sizeof pattern, stdin)){
pattern[strcspn(pattern, "\n")] = 0;//chomp newline
char *p = pattern;
if(invalid(p)){
continue;
}
while(*p){
int n = strtoul(p, &p, 10);
while(n--)
putchar(*p);
++p;
}
puts("");
}
}