Errors using ternary operator in c

血红的双手。 提交于 2019-11-27 05:36:07

In C the ternary operator is defined like

logical-OR-expression ? expression : conditional-expression

where conditional expression is defined like

logical-OR-expression

The assignment operator has a lower priority than the OR operator. Thus you have to write

a >= 5 ? b = 100 : ( b = 200 );

Otherwise the compiler consideres the expression like

( a >= 5 ? b = 100 :  b ) = 200;

As the result of the ternary operator in C is not an lvalue then the above expression is invalid and the compiler issues an error.

From the C Standard:

the result is the value of the second or third operand (whichever is evaluated), converted to the type described below

and footnote:

110) A conditional expression does not yield an lvalue.

Take into account that there is an essential difference between the operator definition in C and C++. In C++ it is defined as

logical-or-expression ? expression : assignment-expression

In C++ the same GCC compiles the code successfully

#include <iostream>

int main() 
{
    int a = 10, b;

    a >= 5 ? b = 100 : b = 200;

    std::cout << "b = " << b << std::endl;

    return 0;
}

you can put it in braces for it to work.. like

(a>=5)?(b=100):(b=200);

and please assign a return type for your function main()

This error is caused by the syntax of the conditional-expression which is

logical-OR-expression ? expression : conditional-expression

Therefore, the part after : must be able to parse b = 200. However, conditional-expression cannot parse that, because an assignment expression has less precedence - you would need to put a parenthesis around the assignment expression

a>=5 ? b=100 : (b=200);

But the fact that you need a parenthesis here does not mean that the expression otherwise is parsed as (a>=5 ? b=100 : b) = 200, it is just a compiler's internal artefact that in the error message it talks about the left operand of assignment. The C language has the following two rules for the assignment expression syntax, and the rule that matches is applied

conditional_expression
unary_expression '=' assignment_expression

This interferes with recursive descent parsers, that would simply invoke parseConditionalExpression, and check what token follows. Therefore some C parser implementations choose to not give a syntax error here, but parse it as though the grammar said conditional_expression '=' ... above, and later when inspecting the parse tree, validate that the left hand side is an lvalue. For example, the Clang source code says

/// Note: we diverge from the C99 grammar when parsing the assignment-expression
/// production.  C99 specifies that the LHS of an assignment operator should be
/// parsed as a unary-expression, but consistency dictates that it be a
/// conditional-expession.  In practice, the important thing here is that the
/// LHS of an assignment has to be an l-value, which productions between
/// unary-expression and conditional-expression don't produce.  Because we want
/// consistency, we parse the LHS as a conditional-expression, then check for
/// l-value-ness in semantic analysis stages.

And the GCC parser's source code says

/* ...
In GNU C we accept any conditional expression on the LHS and
diagnose the invalid lvalue rather than producing a syntax
error. */
user50149

The parentheses should be around the condition in this case.

(a>=5) ? b=100 : b=200; should compile correctly

RE: K&R The C Programming Language (2nd):

Parentheses are not necessary around the first expression of a conditional expression, since the precedence of ?: is very low, just above assignment. (emphasis mine) They are advisable anyway, however, since they make the condition part of the expression easier to see.

Try this one! Because the ternary operator returns the value you have to assgin it to b!

#include <stdio.h>

main() {

    int a = 10, b;
    b = a >= 5 ? 100 : 200;
    printf("%d" , b);

}
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