为什么output说变长可能没有初始化？

Question

 char letM[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
 int nr;
 printf("enter a number between 7 and 15\n");
 scanf("%d", &nr);
 for (int j = 0; j<=nr-5; j++){
   char letter[j] = letM[rand()%26+1];
   printf("%c", letter);
 }

this code should be stamp nr-5 letter but when I run it, output say error: variable-sized object may not be initialized

Answer 1

char letter[j]; defines letter to be an array containing j elements, each of which is a char . Because j is a variable, this is called a variable length array .

char letter[j] = letM[rand()%26+1]; defines such an array and attempts to initialize it with the value of letM[rand()%26+1]; . The C standard does not provide for initializing variable length arrays, and that is why you are getting an error message about it. (They must be given values via ordinary assignments statements or other means, not via initializers in declarations.)

You may have meant to declare letter to be a single char . In this case, change the code to char letter = letM[rand()%26+1]; .

Additionally, letM[rand()%26+1] indexes the array incorrectly. In C, indices start with zero, so you should not add one. Use letM[rand()%26] .

Answer 2

You have to declare what letter is before using it. Here's an option if you want to store the letters. Besides, when you print letter don't forget to print letter[j] instead.

     char letter[21];
     char letM[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
     int nr;
     printf("enter a number between 7 and 15\n");
     scanf("%d", &nr);
     for (int j = 0; j<=nr-5; j++){
       letter[j] = letM[rand()%26+1];
       printf("%c", letter[j]);
     }

Otherwise, as other suggested, here if you just want to print the letter, you don't even need a variable:

 char letM[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
 int nr;
 printf("enter a number between 7 and 15\n");
 scanf("%d", &nr);
 for (int j = 0; j<=nr-5; j++){
   printf("%c", letM[rand()%26+1]);
 }