Increment numbers in char array separated by different delimiters - c++

Question

I have string like this 1-2,4^,14-56
I am expecting output 2-3,5^,15-57
char input[48];
int temp;
char *pch;
pch = strtok(input, "-,^");
while(pch != NULL)
{
char tempch[10];
temp = atoi(pch);
temp++;
itoa(temp, tempch, 10);
memcpy(pch, tempch, strlen(tempch));
pch = strtok(NULL, "-,^");
}
After running through this if I print input it prints only 2 which is first character of the updated string. It does not print all characters in the string. What is the problem with my code?

Answer 1

For plain C, use the library function strtod. Other than atoi, this can update a pointer to the next unparsed character:
long strtol (const char *restrict str, char **restrict endptr, int base);
...
The strtol() function converts the string in str to a long value. [...] If endptr is not NULL, strtol() stores the address of the first invalid character in *endptr.
Since there may be more than one 'not-a-digit' character between the numbers, skip them with the library function isdigit. I placed this at the start of the loop so it would not accidentally convert a string such as -2,3 to -1,4 -- the initial -2 would be picked up first! (And if that is a problem elsewhere, there is also a strtoul.)
Since it appears you want the result in a char string, I use sprintf to copy the output into a buffer, which must be large enough for your possible input plus extra characters caused by a decimal overflow.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <errno.h>
#include <limits.h>
int main (void)
{
char *inputString = "1-2,4^,14-56";
char *next_code_at = inputString;
long result;
char dest[100], *dest_ptr;
printf ("%s\n", inputString);
dest[0] = 0;
dest_ptr = dest;
while (next_code_at && *next_code_at)
{
while (*next_code_at && !(isdigit(*next_code_at)))
{
dest_ptr += sprintf (dest_ptr, "%c", *next_code_at);
next_code_at++;
}
if (*next_code_at)
{
result = strtol (next_code_at, &next_code_at, 10);
if (errno)
{
perror ("strtol failed");
return EXIT_FAILURE;
} else
{
if (result < LONG_MAX)
dest_ptr += sprintf (dest_ptr, "%ld", result+1);
else
{
fprintf (stderr, "number too large!\n");
return EXIT_FAILURE;
}
}
}
}
printf ("%s\n", dest);
return EXIT_SUCCESS;
}
Sample run:
Input: 1-2,4^,14-56
Output: 2-3,5^,15-57

Answer 2

There are two major problems with this code:
First of all,
pch = strtok(input, ",");
When applied to the string 1-2,4^,14-56 will return the token 1-2.
When you call atoi("1-2") you'll get 1, which gets converted to 2.
You can fix this by changing the first strtok to pch = strtok(NULL, "-,^");
Second of all, strtok modifies the string, which means that you lose the original delimiter found. As this looks like a homework exercise, I'll leave you to figure out how to get around this.

Answer 3

I think this could by easier using regular expressions(and C++ instead of C of course):
Complete exmaple:
#include <iostream>
#include <iterator>
#include <regex>
#include <string>
int main()
{
// Your test string.
std::string input("1-2,4^,14-56");
// Regex representing a number.
std::regex number("\\d+");
// Iterators for traversing the test string using the number regex.
auto ri_begin = std::sregex_iterator(input.begin(), input.end(), number);
auto ri_end = std::sregex_iterator();
for (auto i = ri_begin; i != ri_end; ++i)
{
std::smatch match = *i; // Match a number.
int value = std::stoi(match.str()); // Convert that number to integer.
std::string replacement = std::to_string(++value); // Increment 1 and convert to string again.
input.replace(match.position(), match.length(), replacement); // Finally replace.
}
std::cout << input << std::endl;
return 0;
}
Output:
2-3,5^,15-57

Answer 4

strtok modifies the string you pass to it. Either use strchr or something like that to find the delimiters or make a copy of the string to work on.