Changing number from character array to int array with decimal value - c

Question

I am starting to learn C and I am having problem with changing numbers from character array to integer array. Here is my Code and I got random number and I have no idea why I am having those numbers.
The answer is 0 , 45, 45, 45
and I used 15 for plus sign, 25 for minus, 35 for multiply and 45 for division just for testing.
And also the answer which I want is if char[]="+123"-> int[0]=15;int[1]=1;int[2]=2;int[3]=3;
Thanks in advance and very much appreciated!
#include <stdio.h>
int main(){
int i=0;
char retezec[]="123+";
int array_length=(sizeof(retezec) / sizeof(retezec[0])-1);
int new_array[50];
while(retezec[i++]!='\0'){
switch(retezec[i]){
case '0':
new_array[i]=0;
case '1':
new_array[i]=1;
case '2':
new_array[i]=2;
case '3':
new_array[i]=3;
case '4':
new_array[i]=4;
case '5':
new_array[i]=5;
case '6':
new_array[i]=6;
case '7':
new_array[i]=7;
case '8':
new_array[i]=8;
case '9':
new_array[i]=9;
case '+':
new_array[i]=15;
case '-':
new_array[i]=25;
case '*':
new_array[i]=35;
case '/':
new_array[i]=45;
}
}
for(int i=0;i<array_length;i++){
printf("%d\n",new_array[i]);
}
}

Answer 1

Adding with the previous answers the reason you are getting 45 every time is that you haven't given the break keyword in every condition of switch case. Just add it and you will get the desired output :)
For example:
case '0':
new_array[i]=0;
break;
See the link below to have a good understanding. Hope it will help you.
https://www.programiz.com/c-programming/c-switch-case-statement

Answer 2

In your code -
while(retezec[i++]!='\0'){
switch(retezec[i]){
....
i is 0 when while loop condition is first evaluated, but in switch i already becomes 1. So your switch block checks elements from index 1 and not 0. Same way new_array is populated from index 1. The new_array[0] then have garbage value (as you are not initializing new_array)
You can use a for loop which will loop until array_length instead of the while loop like -
for (i = 0; i < array_length; i++) {
switch(retezec[i]) {
....
}
}
Also the switch case looks like an overkill. May be replace it with if-else

Answer 3

This can be boiled down significantly knowing that the C standard define the binary representation of 0 to 9 to necessarily be continuous.
#include <stdlib.h> /* for calloc() */
#include <stdio.h> /* for fprintf() */
int main(void)
{
char retezec[] = "123+";
size_t array_length = sizeof retezec / sizeof *retezec; /* sizeof evaluates to size_t not int. */
int * new_array = calloc(array_length * sizeof *new_array); /* Only get as much as needed, and have set to all 0s. */
if (NULL == new_array)
{
perror("calloc() failed");
return EXIT_FAILURE;
}
for (size_t i = 0; retezec[i] != '\0'; ++i) /* Use a for-loop where a for-loop can be used. */
{
switch(retezec[i])
{
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
new_array[i] = retezec[i] - '0';
break;
case '+':
new_array[i] = 15;
break;
case '-':
new_array[i] = 25;
break;
case '*':
new_array[i] = 35;
break;
case '/':
new_array[i] = 45;
break;
default:
sprintf(stderr, "Unhandled character: '%c'\n", retezec[i]);
break;
}
}
for(size_t i = 0; i < array_length; ++i)
{
printf("%d\n", new_array[i]);
}
free(new_array);
return EXIT_SUCCESS;
}