尝试设置和打印指针的值

Question

I have some code that prompts a user to read in a file that has the format: [name] [someInt], but not all lines have [name]. I thus parse each line into a string array, and if it's length of 2, then it has a name and does a strcmp to find a match and then prints out the int associated. However, I'm running into some issues where I get

error: invalid operands to binary * (have 'char *' and 'char *')

when compiling at the printf("%s\\n" *ans); line

    char * ans = NULL;

//open and read file line by line, below code is in line by line while loop
             char ** res  = NULL;
             char *  p    = strtok (line, " ");
             int n_spaces = 0, i;
             while (p) {
                 res = realloc (res, sizeof (char*) * ++n_spaces);

                 if (res == NULL) {
                     exit (-1); /* memory allocation failed */
                 }

                res[n_spaces-1] = p;
                p = strtok (NULL, " ");
                printf("%d\n", n_spaces);
                if(n_spaces == 2 && (strcmp(name,res[0]) == 0)) {
                    nameMatch = true;
                    printf("MATCH FOUND!\n");
                    ans = res[1];
                    printf("%s\n" *ans);
                    break;
                }
            }

Answer 1

printf("%s\n" *ans);
             ^

You're missing a comma between these arguments. The compiler is interpreting the * as multiplication, and not understanding how you expect it to multiply two strings.

Even with this change, you'll (probably?) still get a warning about types. Remove the * entirely; I'm pretty sure you want to pass the string pointer to printf , not the first character of the string.

Answer 2

you are telling that res is NULL. Then, before malloc you are using realloc? Your mistake might be this.

Answer 3

printf("%s\\n", *ans); instead of printf("%s\\n" *ans);