Declaring a variable gives an error, makes me think there's something else I'm missing

Question

I have a function in C that compiles fine until I add another case to my switch statement and declare a variable.

Here's the function

void evaluate(int num_inputs) {
struct gate *ptr = gatehead;
while (ptr->next != NULL) {
    switch (ptr->kind) {
        case 0:
            if (get_input_value(ptr->params[0]) && get_input_value(ptr->params[1])) {
                write_to_output(true, ptr->output);
            } else {
                write_to_output(false, ptr->output);
            }
            break;
        case 1:
            if (get_input_value(ptr->params[0]) || get_input_value(ptr->params[1])) {
                write_to_output(true, ptr->output);
            } else {
                write_to_output(false, ptr->output);
            }
            break;
        case 2:
            if (!(get_input_value(ptr->params[0]) && get_input_value(ptr->params[1]))) {
                write_to_output(true, ptr->output);
            } else {
                write_to_output(false, ptr->output);
            }
            break;
        case 3:
            if (!(get_input_value(ptr->params[0])) && !(get_input_value(ptr->params[1]))) {
                write_to_output(true, ptr->output);
            } else {
                write_to_output(false, ptr->output);
            }
            break;
        case 4:
            if ((get_input_value(ptr->params[0]) || get_input_value(ptr->params[1])) && !(get_input_value(ptr->params[0]) && get_input_value(ptr->params[1]))) {
                write_to_output(true, ptr->output);
            } else {
                write_to_output(false, ptr->output);
            }
            break;
        case 5:
            if (get_input_value(ptr->params[0])) {
                write_to_output(false, ptr->output);
            } else {
                write_to_output(true, ptr->output);
            }
            break;
        case 6:
            if (get_input_value(ptr->params[0])) {
                write_to_output(true, ptr->output);
            } else {
                write_to_output(false, ptr->output);
            }
            break;
        case 7:
            char* d = "DECODER";
            printf("%s", d);
            break;
    }
    ptr = ptr->next;
}

}

If I comment out case 7, everything works fine. Even keeping the print and break compiles, but declaring the char* causes the error "expected expression"

Answer 1

The problem is that the case 7: label is immediately followed by a variable declaration.

    case 7:
        char* d = "DECODER";

A label can only be applied to a statement, and a declaration is not considered a statement. You can get around this by adding an empty statement before the case label:

    case 7:
        ;
        char* d = "DECODER";