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application of stacks.
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i have this question..
show the state of the stack and the value of each variable after execution of the following statements.
let A=5, B=3, C=7
push A onto stack
push C*C onto stack
pop item fron the stack and store in B
push B+A onto stack
pop item from the stack and store in A
pop item for the stacj and sotore in B
-------------------------------------
what will be the value of B after C*C is popped out of the stack?? will the predefined value of B be overwritten?? and what if B is pushed onto the stack, will the value of the pushed B be the value of the result of C*C??
for example..
this???
push A onto stack = 5
push C*C onto stack = 49
pop item fron the stack and store in B = B=49
push B+A onto stack = 49+5=54
pop item from the stack and store in A = A=54
pop item for the stacj and sotore in B = B=5
??????????????????????
or this??
push A onto stack = 5
push C*C onto stack = 7*7=49
pop item fron the stack and store in B = B=49
push B+A onto stack = 3+5=8
pop item from the stack and store in A = A=8
pop item for the stacj and sotore in B = B=5
???????????????????/
HELP
show the state of the stack and the value of each variable after execution of the following statements.
let A=5, B=3, C=7
push A onto stack
push C*C onto stack
pop item fron the stack and store in B
push B+A onto stack
pop item from the stack and store in A
pop item for the stacj and sotore in B
-------------------------------------
what will be the value of B after C*C is popped out of the stack?? will the predefined value of B be overwritten?? and what if B is pushed onto the stack, will the value of the pushed B be the value of the result of C*C??
for example..
this???
push A onto stack = 5
push C*C onto stack = 49
pop item fron the stack and store in B = B=49
push B+A onto stack = 49+5=54
pop item from the stack and store in A = A=54
pop item for the stacj and sotore in B = B=5
??????????????????????
or this??
push A onto stack = 5
push C*C onto stack = 7*7=49
pop item fron the stack and store in B = B=49
push B+A onto stack = 3+5=8
pop item from the stack and store in A = A=8
pop item for the stacj and sotore in B = B=5
???????????????????/
HELP
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Join Date: Jun 2005
Posts: 2,044
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Solved Threads: 139
When you store something in B, that means the predefined value will be overwritten.
They want you to show the state of the stack, ie what's in it.
So more like
push A onto stack [5]
push C*C onto stack [49,5]
...
So more like
push A onto stack [5]
push C*C onto stack [49,5]
...
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•
Join Date: Jun 2005
Posts: 2,044
Reputation:
Solved Threads: 139
Anyway, your first example (where A gets assigned 54) is perfectly right.
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Join Date: Sep 2009
Posts: 4
Reputation:
Solved Threads: 0
Newbie Poster
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Originally Posted by scias23 
i have this question..
show the state of the stack and the value of each variable after execution of the following statements.
let A=5, B=3, C=7
push A onto stack
push C*C onto stack
pop item fron the stack and store in B
push B+A onto stack
pop item from the stack and store in A
pop item for the stacj and sotore in B
-------------------------------------
what will be the value of B after C*C is popped out of the stack?? will the predefined value of B be overwritten?? and what if B is pushed onto the stack, will the value of the pushed B be the value of the result of C*C??
for example..
this???
push A onto stack = 5
push C*C onto stack = 49
pop item fron the stack and store in B = B=49
push B+A onto stack = 49+5=54
pop item from the stack and store in A = A=54
pop item for the stack and store in B = B=5
??????????????????????
or this??
push A onto stack = 5
push C*C onto stack = 7*7=49
pop item fron the stack and store in B = B=49
push B+A onto stack = 3+5=8
pop item from the stack and store in A = A=8
pop item for the stacj and sotore in B = B=5
???????????????????/
HELP
your first example
ie
push A onto stack = 5
push C*C onto stack = 49
pop item fron the stack and store in B = B=49
push B+A onto stack = 49+5=54
pop item from the stack and store in A = A=54
pop item for the stack and store in B = B=5
is definetly right..ok
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