CODEWITHSUNDEEP
c
SOUFYANE OUCHAALI
SOUFYANE OUCHAALI

Reputation: 21

How does this type of loop work: `for (; x++;);`

How does this type of loop work and why is it printing x=1; instead x=3 ?

#include <stdio.h>
#include <stdlib.h>

int main() {
    char x = 2;
    for (; x++;);
    printf("x=%d", x);
}

Upvotes: 1

Views: 756

Answers (4)

abelenky
abelenky

Reputation: 64710

for(;x++;);

This is a loop with no body.
It will stop when x++ evaluates to False (0).
Because x starts at 2, and has type char, it will (most likely) increment to +127, then wrap around to -128, then increment to up to 0, when the loop finally ends.

(As the commenters point out, wrap-around on signed types is not part of the spec, and is technically implementation-defined behavior. But most common implementations do wrap around in a predictable fashion.)

When the loop is finally done, the value of x is 1, due to the post-increment, and the value will be printed out.

The final output will be: x=1

Upvotes: 0

Eric Postpischil
Eric Postpischil

Reputation: 223663

After Int is corrected to int, the loop for(;x++;); will execute the null statement ; until x++ evaluates as false (zero).

Since x starts at two, it is incremented to three, then four, and so on. This continues until x reaches the maximum value of a char. Then adding one yields a value that char cannot represent.1 This number is then converted to char for storing in x.

If char is unsigned, this conversion wraps to zero, so zero is stored in x. If char is signed, the result of the conversion is implementation-defined (or a signal is raised). A typical result is to wrap to the minimum char value, commonly −128. In this case, the loop then proceeds to increment x through −127, −126, and so on, until it reaches zero.

Thus, in either of the common cases above, x eventually reaches zero. Then the next evaluation of x++ increments x to one and evaluates to zero. Then the loop exits, and printf("x=%d",x); prints “x=1”.

Other behaviors are possible. For example, a C implementation could define the conversion of an out-of-range value to char to produce 127, and then the loop would not terminate. (Interestingly, though, the rule in C 2018 6.8.5 6 allows a compiler to assume the loop terminates, even though actually executing it would not.)

Footnote

1 The arithmetic is done in the int type, so it does not overflow (except in hypothetical bizarre C implementations where char is the same width as int). In specifying the postfix ++ operator, C 2018 6.5.2.4 2 refers to the additive and assignment operators, which promote char operands to int.

Upvotes: 9

glglgl
glglgl

Reputation: 91119

This code relies on overflow behaviour, and I think with signed types (which char possibly can be) this is undefined behaviour.

But if it isn't, the following works:

char x=2;

x gets assigned 2.

for(;x++;);

This is a loop with an empty method body and its break condition x++. You could as well write while (x++);.

It loops, doing nothing but incrementing x, until x evaluates "false" (0) before incrementing:

  • It gets the value (2) and increments. This value is != 0, so repeat.
  • It gets the value (3) and increments. This value is != 0, so repeat.
  • It gets the value (4) and increments. This value is != 0, so repeat.
  • ...
  • It gets the value (127) and increments. This value is != 0, so repeat.
  • It gets the value (128 if unsigned, -128 if signed AND the overflow behaviour is "two-s complement") and increments. This value is != 0, so repeat.
  • It gets the value (255 if unsigned, -1 if signed) and increments. This value is != 0, so repeat.
  • It gets the value (0) and increments. The value is == 0, so break the loop.

That last increment makes x have the value 1.

Thus,

printf("x=%d",x);

prints x=1.

Upvotes: 0

Sai Suman Chitturi
Sai Suman Chitturi

Reputation: 422

char uses one byte of memory. So, its range is from -128 to 127.

char x=2;
for(;x++;);

This loop will run for values x = {2, 3, ..., 127, -128, -127, ..., -1, 0}.

As soon as x becomes 0, the condition check will stop the loop but the increment will also take place.

Hence, 0 becomes 1 and hence 1 will be printed.

Upvotes: -1

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