![]() In this case it works because the handler just return the current call depth. Note that this technique requires some care, because there must be enough stack space for the handler to work. In the solution the function Recursion calls itself or else catches Storage_Error indicating stack overflow. Storage_Error can be handled without termination of the task. In particular it is propagated upon stack overflow within the task where this occurs. In Ada Storage_Error exception is propagated when there is no free memory to accomplish the requested action. Note that unlike some solutions in other languages this one does not crash (though usefulness of this task is doubtful). Put_Line ( "Recursion depth on this system is" & Integer'Image (Recursion ( 1 ) ) ) Text_IO įunction Recursion (Depth : Positive ) return Positive is This means anything from this place up to SP is free for the The program does not need any memory beyond this point. The number of free stack words is now in HL, output it Ora h so HL should be divided by two to give the actual ![]() Xra a This gives bytes, but a call takes two bytes Lxi h,-top Subtract highest used location from stack pointer ![]() Of calls you can make before the stack is full (and would overwrite your program). (addresses are 16 bits), if you then divide it by 2, you get the maximum amount Program is actually using, and subtract this from SP, you get theĪmount of free stack memory, in bytes. Therefore, if you take the address of the highest byte of memory that your Location of the topmost stack item (or, if the stack is considered 'empty', On the SP (stack pointer) register, which always points to the The 8080 processor decides where the stack is depending If all you need to know is how much room there is beforehand, it's possible to (Value will differ depending on system, CP/M version, memory size, etc.) GUARD: equ $+2 Make sure it is not a valid return address If the guard is overwritten, the stack has overflowed. so all memory beyond here is free for use by the stack. The program doesn't need any memory after this location, Mov a,h If so, the answer should be zeroĬz recur If it is, do another recursive call Lhld guard (subtract the original value from the Lxi d,-GUARD See if the guard is intact (stack not full) ![]() can make, and the stack is back to the beginning. ěC now holds the maximum amount of recursive calls one This way, a stack overflow can be caughtĪt run-time, at the cost of some overhead per call. Something important, so in that case, the stack is full. If it isn't, the next call will overwrite When you doĪ recursive call, you first check if it is still intact. Store a word of data just beyond the last byte of memory that the program wants to use. One way of doing this is by using a stack guard (also known as a stack sentinel). (Note that they will give slightly different answers on the same system,Īs one program is slightly bigger than the other, leaving a little less room for the stack.) Below are two ways of finding out the maximum recursion limit Simply overwrite other data or code, and the program will crash. However, it has no support for any kind of memory protection. Which is still true on modern Intel processors. It also set the convention that the machine stack growsĭownward into memory (i.e., the topmost item on the stack is one word below the second one, etc.), Recursive calls could be somewhat practical. Had an on-die call stack that was limited to 7 levels.) This means it was the first one on which The 8080 was the first Intel processor to support a stack in RAM. Lda $0100 if this no longer equals $BE the stack has overflowedīne StackHasOverflowed 8080 Assembly Pick a value that is unlikely to be used in your program. Unlike the 8086 and Z80, which require all pushes/pops to be exactly two bytes, the 6502's stack will likely contain both 1 byte registers and 2 byte return addresses. Reading the current stack pointer is unreliable, as there is no requirement that the stack be "aligned" in any way. Since each JSR pushes a 2-byte return address onto the stack, the hardware limit of recursion is 128 calls. If the stack were to reach $01, and another JSR is executed, the stack would underflow to $FE, and the bottom of the stack would get clobbered. The stack pointer is an 8-bit value, and assumes that the stack is always between $0100-$01FF. First, it is fixed at the $0100-$01FF address range. The 6502's hardware stack isn't like other processors. 22.1 A more 'canonical' way of doing it.
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