将修改后的strtok令牌转换回原始字符串的“语法”

Question

Is there a way after you've split a string into tokens to convert these tokens back into the original string (with punctuation, and original case-sensitivity)?

I'm fairly certain strcat will be invoked, but what would be a good method to maintain the original string's punctuation & case-sensitivity?

Assume that I need to modify each and every token individually before concatenating them back into the original string's structure.

Edit: The title has been reworded & clarified, I hope.

char* msg = malloc(10000);
char* msg2 = malloc(10000);
char* buffer;
char buffer2[10000];

printf("input: \n");
fgets(msg, 9999, stdin);

msg2 = strdup(msg);
buffer = strtok(msg2, " ;,.!?\n");
Node* ptr; // assume this is initialized to a linked list containing 2 strings per node (str, str2)
while (buffer) {

        for (int i = 0; i < strlen(buffer); i++) { //strcmp is case-sensitive
            if (isalpha(buffer[i])) {
                buffer[i] = tolower(buffer[i]);
            }
        }
        while (ptr) {
            if ((ptr != NULL) && (strcmp(ptr->str, ptr->str2) != 0)) {
                buffer = ptr->str2; // modifies the token 
                break;
            }
    }
    ptr = ptr->next;
    }
    strcat(buffer2, " "); // this will concatenate the modified tokens back together with spaces
    strcat(buffer2, buffer); // but I lose the original string's structure (i.e.: punctuation, case-sensitivity)
    buffer = strtok(NULL, " ;,.!?\n");

    printf("%s", buffer2); // so how should I get the original string's structure back with the modified tokens?

Answer 1

If you want to use strtok() and want to access the original string later, please keep a copy of the original string, since strtok() will destroy the original string.

// we have char *st ;

char *tem = malloc(strlen(st)+1);
strcpy(tem, st);

Now you can access tem as original string later.