Error in C code error: expression is not assignable

Question

I'm writing a function that is suppose to print out the description of a programs execution. A function in my program uses 0 as a signal for a base-10 numeric conversion.

I would like my program to have friendly output, and tell the user if a number has been converted to base 10, instead of letting the program say the number was converted from base 0.

When I attempt to compile this code, I get an error message which says 'expression is not assignable'.

I'm compiling on command line with cc compiler

Apple LLVM version 7.3.0 (clang-703.0.29)

Any idea what this error means and how to correct? Thanks.

void foo( int base ){

    int theBase;

    base == 0 ? theBase = 10: theBase = base;

    printf("%s%d\n", "The base is ", theBase)
}

error message:

error: expression is not assignable base == 0 ? theBase = 10: theBase = base; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ^

Answer 1

What you are doing here is a conditionnal assignation.

Usually you can do it like that:

if (base == 0)
    theBase = 10;
else
    theBase = base;

Here you chosed to use a ternary expression. It does work a bit like an if/else structure, but it is really different.

The ternary return a value, it's not made to execute code based on a condition. No, it return a value based on a condition.

So here, you have to do:

theBase = (base == 0 ? 10 : base);

(the parenthesis are not required, but it's a lot better to avoid error).

In fact, you can make a ternary execute code, in a multiple way, like returning a function:

int my_function()
{
    printf("Code executed\n");
    return (10);
}

/* ... */

theBase = (base == 0 ? my_function() : base);

EDIT:

And yes you can use that code:

base == 0 ? (theBase = 10) : (theBase = base);

But it's pretty useless to use a ternary in that case, because you still have to duplicate the theBase = X code.