char数组中由不同分隔符分隔的递增数字

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

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 2

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 3

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.

Answer 4

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