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This page (revision-72) was last changed on 01-May-2018 03:30 by Roland B. Wassenberg  

This page was created on 20-Feb-2010 21:56 by Carsten Strotmann

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At line 7 removed 4 lines
\\
A = 10 + 20 * B\\
\\
To perform this line, the interpreter has to read the entire line, look up the value of B (let's say 30), realize that the * has to be performed before +, and then finally convert those into instructions something like:\\
At line 12 changed 4 lines
get(B,temp1)\\
mul(20,temp1,temp2)\\
add(10,temp2,temp3)\\
put(temp3,A)\\
{{A = 10 + 20 * B}}\\
At line 17 changed one line
In contrast, in a stack-based system, the programmer organizes the code in the fashion it will ultimately be performed. The equivalent would be:\\
To perform this line, the interpreter has to read the entire line, look up the value of B (let's say 30), realize that the * has to be performed before + and order the instructions correctly, and then finally convert those into instructions something like:\\
\\
{{get(B,temp1)}} - get the value in B and store it in temp1\\
{{multiply(20,temp1,temp2)}} - multiply that value by 20 and store the result in temp2\\
{{add(10,temp2,temp3)}} - add 10 to temp2 and store the result in temp3 \\
{{put(temp3,A)}} - store the value of temp3 into the variable A\\
\\
In contrast, in a stack-based system, the programmer organizes the code in the fashion it will ultimately be performed. The equivalent would be something like:\\
At line 19 changed 2 lines
B 20 mul\\
10 add\\
{{B 20 mul}}\\
{{10 add}}\\
At line 22 changed one line
When this code is performed, the interpreter pushes the value of B on the stack, then 20. It then encounters the mul, which removes the last two items, the 30 and 20, multiplies them, and puts the result back on the stack. Next, it pushes 10 on the stack, and encounters add, taking the two values, adding them, and putting the result back on the stack.
When this code is performed, the interpreter pushes the value of B on the stack, then 20. It then encounters the mul, which removes the last two items, the 30 and 20, multiplies them, and puts the result back on the stack. Next, it pushes 10 on the stack, leaving the top two locations containing 60 and 10. It then encounters add, taking the two values, adding them, and putting the result back on the stack. The top of the stack now contains the result, 70.
At line 24 changed one line
Notice that the stack-based version *has no temporary values*, and only reads a single instruction at a time, not an entire line of code. As a result, the parser is much simpler, smaller and requires less memory to run. This, in turn, generally makes it much faster, comparable to compiled programs.
Notice that the stack-based version ''has no temporary values'', and only reads a single instruction at a time, not an entire line of code. As a result, the parser is much simpler, smaller and requires less memory to run. This, in turn, generally makes it much faster, comparable to compiled programs.
At line 108 removed one line
Version Date Modified Size Author Changes ... Change note
72 01-May-2018 03:30 6.695 kB Roland B. Wassenberg to previous
71 28-Apr-2018 15:44 6.695 kB Maury Markowitz to previous | to last
70 28-Apr-2018 15:38 6.295 kB Maury Markowitz to previous | to last
69 28-Apr-2018 15:36 6.27 kB Maury Markowitz to previous | to last
68 26-Apr-2018 22:31 4.923 kB Maury Markowitz to previous | to last
67 26-Apr-2018 22:19 4.336 kB Maury Markowitz to previous | to last
66 29-Aug-2017 04:50 3.271 kB Roland B. Wassenberg to previous | to last
65 25-Feb-2017 16:55 3.292 kB Carsten Strotmann to previous | to last
64 05-Jul-2016 10:56 3.153 kB Roland B. Wassenberg to previous | to last
63 02-Jul-2016 01:07 3.103 kB Roland B. Wassenberg to previous | to last
62 01-Jul-2016 18:40 3.062 kB Roland B. Wassenberg to previous | to last
61 23-Jan-2014 18:28 3.048 kB Carsten Strotmann to previous | to last
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