如何使用 scanf 从缓冲区中提取一些格式化的字符串？

Question

I need to extract both "rudolf" and "12" from that long string: "hello, i know that rudolph=12 but it so small..." using scanf , how can I do it?

This buffer can contains any formatted strings like ruby=45 or bomb=1 , and I dont know it in advance.

I am trying something like that, but it was unsuccessful

#include <stdio.h>

int main()
{
    char sentence[] = "hello, i know that rudolph=12 but it so small...";
    char name[32];
    int value;

    sscanf(sentence, "%[a-z]=%d", name, &value);
    printf("%s -> %d\n", name, value);

    getchar();
    return 0;
}

Answer 1

Iterate through the sentence using a temporary pointer and %n to extract each sub-string.
%n will give the number of characters processed by the scan to that point. Add that to the temporary pointer to advance through the sentence.
Try to parse from each sub-string the name and value. The scanset %31[^=] will scan a maximum of 31 characters, leaving room in name for a terminating zero. It will scan all characters that are not an = . Then the format string will scan the = and try to scan an integer.

#include <stdio.h>
#include <stdlib.h>

int main (void) {
    char sentence[] = "hello, i know that rudolph=12 but it so small...";
    char string[sizeof sentence] = "";
    char name[32] = "";
    char *temp = sentence;
    int value = 0;
    int count = 0;
    int parsed = 0;

    while (1 == sscanf(temp, "%s%n", string, &count)) {
        temp += count;
        if (2 == sscanf(string, "%31[^=]=%d", name, &value)) {
            parsed = 1;
            break;
        }
    }
    if (parsed) {
        printf("%s %d\n", name, value);
    }

    return 0;
}

Answer 2

You can write your own string serach engine. Which could be quite simple, let's for example:

• advance until you find one of [az]
  • remember position
  • advance until the end of [az]
  • check if it's = now
    • if it is, there was our variable name
    • advance until end of value
    • return it
• if there's no = , omit everything that is not a [az] ie. can't be variable name

A sample program:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>

struct find_assignment_s {
    const char *line;
    const char *var;
    size_t varlen;
    const char *val;
    size_t vallen;
};

struct find_assignment_s find_assignment_init(const char *line) {
    return (struct find_assignment_s){.line = line};
}

int find_assignment(struct find_assignment_s *t) {
    while (*t->line) {
        const char *p = t->line;
        while (*p && isalpha((unsigned char)*p)) p++;
        // There is at least one alphabetic character and there is a space right after.
        if (t->line != p && *p == '=') {
            // Found a "variable="!
            t->var = t->line;
            t->varlen = p - t->line;
            // value is up until a space is found
            t->val = p + 1;
            while (*p && !isspace((unsigned char)*p)) p++;
            t->vallen = p - t->val;
            // Advance the pointer behind value.
            t->line = *p ? p + 1 : p;
            return 1;
        }
        // Ignore spaces
        while (*p && !isalpha((unsigned char)*p)) p++;
        // Advance over whole word.
        t->line = p;
    }
    return 0;
}

int main() {
    const char line[] =  "hello, i know that rudolph=12 but it so small... a=b c=d fdnajn=123";
    for (struct find_assignment_s fs = find_assignment_init(line);
            find_assignment(&fs) == 1; ) {
        printf("%.*s = %.*s\n", (int)fs.varlen, fs.var, (int)fs.vallen, fs.val);
    }
}

outputs:

rudolph = 12
a = b
c = d
fdnajn = 123