c - 将字符串中的 1-3 等速记符号扩展为 C 中的字符串数组

Question

假设我有一个字符串 XYZ1-3。

我想将其转换为字符串数组。XYZ1,XYZ2,XYZ3.

在 C 中是否有一种优雅的方式来做到这一点？

Answer 1

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

char **expand(const char *string, int *num){//num : out var
    char *id = strdup(string);
    int start, end, len;
    sscanf(string, "%*[A-Z]%n%d-%d", &len, &start, &end);
    id[len] = '\0';
    *num = end-start+1;
    char **array = malloc(*num * sizeof(char*));
    for(int i=0;i < *num ;++i){
        len = snprintf(NULL, 0, "%s%d", id, start + i);
        array[i] = malloc(++len);
        sprintf(array[i], "%s%d", id, start + i);
    }
    free(id);
    return array;
}

int main(){
    int n;
    char **array = expand("XYZ1-3", &n);
    for(int i=0;i<n;++i){
        printf("%s\n", array[i]);
        free(array[i]);
    }
    free(array);
    return 0;
}

---

允许非字母(非 A-Z)id 部分

#include <ctype.h>

int id_length(const char *string){
    //return length of id part.
    int i, len;
    for(i=0;string[i];++i);
    if(i==0)return 0;
    for(i=i-1;isdigit(string[i]) && i>=0;--i);
    if(string[i]!='-') return 0;//bad format
    for(i=i-1;isdigit(string[i]) && i>=0;--i);
    return i+1;
}

char **expand(const char *string, int *num){//num : out var
    char *id = strdup(string);
    int start, end, len;
    len = id_length(string);
    sscanf(string+len, "%d-%d", &start, &end);
    id[len] = '\0';
    *num = end-start+1;
    char **array = malloc(*num * sizeof(char*));
    for(int i=0;i < *num ;++i){
        len = snprintf(NULL, 0, "%s%d", id, start + i);
        array[i] = malloc(++len);
        sprintf(array[i], "%s%d", id, start + i);
    }
    free(id);
    return array;
}