[zedshaw]

Alright, I did finally get this working. Pretty fucking awesome, I had not thought of that. Here's a version everyone can try:

https://gist.github.com/zedshaw/64b3fb6b7ed653852619

I officially concede that because you can work two pointers on a computer to overwrite another location of memory to alter a for-loop (incidentally, there's not UB listed in ANSI for 'alter the variable of a for-loop') that everyone should go back to writing their C code just as K&R intended.

Please, you all should rely on only the '\0' byte terminator of all strings, don't do any bounds checking, don't check the return code of functions, and you will be totally safe.

Because, UB means "I ain't gotta fix it."

Enjoy, now I'm going painting.

[mikeash]

Your insistence that undefined behavior is not at play here is bizarre. It is not possible to construct a pointer into another stack frame like my code does without invoking undefined behavior. To wit:

"When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i−n-th elements of the array object, provided they exist. Moreover, if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q)-1 points to the last element of the array object. If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined."

This is a long-winded standards-language way of saying that if you compute x+y, where x is an array and y is not a valid index in the array or one past the end of the array, the behavior is undefined. The moment I computed 'output', I hit UB. Everything that happens after that is up to the whims of the platform.

[spion]

Things like this make me wonder why is anyone using C at all these days. Rust can't come soon enough.

[Dylan16807]

Also everything that happens before it!