什么函数是从 C 中的字符串中替换子字符串? [英] What function is to replace a substring from a string in C?

问题描述

给定一个 (char *) 字符串，我想查找所有出现的子字符串并将它们替换为备用字符串.我在 <string.h> 中看不到任何实现此目的的简单函数.

Given a (char *) string, I want to find all occurrences of a substring and replace them with an alternate string. I do not see any simple function that achieves this in <string.h>.

推荐答案

优化器应该消除大部分局部变量.tmp 指针用于确保 strcpy 不必遍历字符串来查找空值.tmp 指向每次调用后结果的结尾.(请参阅 Shlemiel thepainter's algorithm 了解 strcpy 为何令人讨厌.)

The optimizer should eliminate most of the local variables. The tmp pointer is there to make sure strcpy doesn't have to walk the string to find the null. tmp points to the end of result after each call. (See Shlemiel the painter's algorithm for why strcpy can be annoying.)

// You must free the result if result is non-NULL.
char *str_replace(char *orig, char *rep, char *with) {
    char *result; // the return string
    char *ins;    // the next insert point
    char *tmp;    // varies
    int len_rep;  // length of rep (the string to remove)
    int len_with; // length of with (the string to replace rep with)
    int len_front; // distance between rep and end of last rep
    int count;    // number of replacements

    // sanity checks and initialization
    if (!orig || !rep)
        return NULL;
    len_rep = strlen(rep);
    if (len_rep == 0)
        return NULL; // empty rep causes infinite loop during count
    if (!with)
        with = "";
    len_with = strlen(with);

    // count the number of replacements needed
    ins = orig;
    for (count = 0; tmp = strstr(ins, rep); ++count) {
        ins = tmp + len_rep;
    }

    tmp = result = malloc(strlen(orig) + (len_with - len_rep) * count + 1);

    if (!result)
        return NULL;

    // first time through the loop, all the variable are set correctly
    // from here on,
    //    tmp points to the end of the result string
    //    ins points to the next occurrence of rep in orig
    //    orig points to the remainder of orig after "end of rep"
    while (count--) {
        ins = strstr(orig, rep);
        len_front = ins - orig;
        tmp = strncpy(tmp, orig, len_front) + len_front;
        tmp = strcpy(tmp, with) + len_with;
        orig += len_front + len_rep; // move to next "end of rep"
    }
    strcpy(tmp, orig);
    return result;
}

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