C:传递和返回指向char*的指针

#include <stdio.h>

char* concatenateStrings(const char* str1, const char* str2) {
    int len1 = 0, len2 = 0;
    while (str1[len1] != '\0') {
        len1++;
    }
    while (str2[len2] != '\0') {
        len2++;
    }
    
    char* result = malloc((len1 + len2 + 1) * sizeof(char));
    int i, j;
    for (i = 0; i < len1; i++) {
        result[i] = str1[i];
    }
    for (j = 0; j < len2; j++) {
        result[i + j] = str2[j];
    }
    result[i + j] = '\0';
    
    return result;
}

int main() {
    const char* str1 = "Hello";
    const char* str2 = " World!";
    
    char* concatenated = concatenateStrings(str1, str2);
    printf("Concatenated string: %s\n", concatenated);
    
    free(concatenated); // Remember to free the dynamically allocated memory
    
    return 0;
}