How to transform, for example, "0" to "ZERO" in C? What problem in this code?

Question

I have to do a program for college whose main objective is another, but in one of the stages of the program, I need to pass two integer random numbers from 0 to 9 for their extensive written formats, in string. I wrote some excerpts from a code that would perform this, but I wasn’t very successful.

#include <time.h>
#include <stdlib.h>
#include <stdio.h>


int main(){
    int i=0, rand1, rand2;
    char num1[7], num2[7];

    printf("Bem vindo! Pronto pra começar? Vamos la!");

    srand(time(NULL));

    while (i<10){
          rand1=rand()%10;
          rand2=rand()%10;

           switch (rand1){
            case 0:
                 num1 = "ZERO";
            case 1:
                 num1 = "UM";
            case 2:
                 num1 = "DOIS";
            case 3:
                 num1 = "TRES";
            case 4:
                 num1 = "QUATRO"
            case 5:
                 num1 = "CINCO";
            case 6:
                 num1 = "SEIS";
            case 7:
                 num1 = "SETE";
            case 8:
                 num1 = "OITO";
            case 9:
                 num1 = "NOVE";
            }
          switch (rand2){
            case 0:
                 num2 = "ZERO";
            case 1:
                 num2 = "UM";
            case 2:
                 num2 = "DOIS";
            case 3:
                 num2 = "TRES";
            case 4:
                 num2 = "QUATRO"
            case 5:
                 num2 = "CINCO";
            case 6:
                 num2 = "SEIS";
            case 7:
                 num2 = "SETE";
            case 8:
                 num2 = "OITO";
            case 9:
                 num2 = "NOVE";
            }


          printf("Quanto e %s vezes %s?: ", num1, num2);

          i++;
          }
    system("pause");
    return 0;


}

Error occurs in string assignment num1 and num2 with the message

incompatible types in assignement of 'const char [5]' to '[char [7]'

Answer 1

You stated char num1[7]. It is an array with 7 characters. But in C there is no array assignment operation. So num1 = "ZERO" doesn’t work.

1. Can transform num1 and a pointer to char only, so when do num1 = "ZERO" it will point to the static chars array. I recommend so since you do not intend to change the data in any way.

2. Uses the function strcpy to copy the string to the array: strcpy(num1, "ZERO");. It works for what you want, but the copy is unnecessary since you will not change the data. A mere pointer suffices.

---

Unrelated, but a better way to write your code would be as follows:

const char* words[10] = {"ZERO", "UM", "DOIS", "TRES", "QUATRO", "CINCO",
                         "SEIS", "SETE", "OITO", "NOVE"};

char* num1 = words[rand1];

Answer 2

It is not possible to assign values to arrays. You need to assign array elements one by one. The C language has some shortcuts to facilitate this assignment when it comes to strings.

num1[0] = 'Z';
num1[1] = 'E';
num1[2] = 'R';
num1[3] = 'O';
num1[4] = '\0';

or, using the shortcut

strcpy(num1, "ZERO"); // lembra-te de incluir o header <string.h>

Answer 3

About build error already have good answers: you cannot assign values like this in C. int and char[7] with a text representing the int are very different things and there is no direct conversion.

And the case is wrong: missing the break;

Some time later, I will suggest the potentially much faster and much simpler

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    const char* numero[] =
    {
        "ZERO", "UM", "DOIS", "TRES", "QUATRO",
        "CINCO", "SEIS", "SETE", "OITO", "NOVE"
    };
    char num1[7] = {0};
    char num2[7] = {0};
    printf("Bem vindo! Pronto pra começar? Vamos la!\n");
    srand(210821);
    int total = 10;
    for(int i=1;i<=total;i+=1)
    {
        int rand1 = rand() % 100;
        strcpy(num1, numero[rand1 / 10]);
        strcpy(num2, numero[rand1 % 10]);
        printf("[%2d/%2d]  %s   e   %s\n", i, total, num1, num2);
    }
    return 0;
}

Which shows

Bem vindo! Pronto pra comeþar? Vamos la!
[ 1/10]  SEIS   e   TRES
[ 2/10]  SEIS   e   SEIS
[ 3/10]  UM   e   OITO
[ 4/10]  QUATRO   e   SEIS
[ 5/10]  ZERO   e   ZERO
[ 6/10]  TRES   e   DOIS
[ 7/10]  SEIS   e   SEIS
[ 8/10]  UM   e   ZERO
[ 9/10]  UM   e   OITO
[10/10]  SEIS   e   QUATRO

With no switch and a single call to rand().

Answer 4

In the C language you cannot assign a string to a vector simply using the assignment operator '='. This serial possible if using the C string type++.

However, there is a set of functions in C included in the header file among which I can quote strcpy that copies a string to a char vector. So in your comparison, if it’s number six, you’d do:

char str[256];
strcpy(str, "SEIS");

And you would now have the string saved in your vector. It copies the right string into its left vector provided it is large enough.