Semantics of `printf(“…”) || printf(“…”) || printf(“…”)`

Question

I\'m wondering what the following statement will print in C?

printf(\"hello\\n\") || (printf(\"goodbye\\n\") || printf(\"world\\n\"));

I\'m usu

Answer 1

First, cout is a C++ invention, never made it back to C, and never will.

Next, printf returns the number of printed characters, so the first call returns non-zero.

As || is short-circuiting boolean-or, none of the following printf-calls will be done.

(| is bitwise-or, and thus not short-circuiting. Added because you are talking about single pipes and @Leeor linked such a question.)

Endresult: hello\n is printed: 5 characters+newline (will be translated, as stdin is text-mode (identity-transformation on Unixoids)).

7.21.6.3 The printf function

Synopsis

#include 
int printf(const char * restrict format, ...);

Description
2 The printf function is equivalent to fprintf with the argument stdout interposed before the arguments to printf.
Returns
3 The printf function returns the number of characters transmitted, or a negative value if an output or encoding error occurred.

6.5.12 Bitwise inclusive OR operator

Synopsis
[...]
Constraints
2 Each of the operands shall have integer type.
Semantics
3 The usual arithmetic conversions are performed on the operands.
4 The result of the | operator is the bitwise inclusive OR of the operands (that is, each bit in the result is set if and only if at least one of the corresponding bits in the converted operands is set).

6.5.14 Logical OR operator

Synopsis
[...]
Constraints
2 Each of the operands shall have scalar type.
Semantics
3 The || operator shall yield 1 if either of its operands compare unequal to 0; otherwise, it yields 0. The result has type int.
4 Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares unequal to 0, the second operand is not evaluated.