Contributors: Nikolaos Giannopoulos, Dimitrios Andronikou
Spring Semester 2021 - 2022
The code based on the main.c file written was based on the ASCII table to allow us to ignore lowercase Latin ,symbols.
Figure 1 shows the ASCII table which in this paper/lab we used as a first check if the input to our routine by taking the 1byte in register r4 is NUL i.e. an empty alphanumeric if it is true then we go to the EXIT label and terminate our routine. Then we check if it is less than the decimal equivalent of / if true then we go to label LOOP and get the next byte otherwise we then check if it is less than the character : if it is then we go to label START otherwise if it is not less we go to the next check and see if it is greater than the character @ if true then we go to label START otherwise again we go to label LOOP and get the next byte i.e. we ensure that if it is a symbol : ? < = > ? @ then we ignore it and go to the next one. When we are inside after label START then we check if the character is greater than the symbol [ if true then we go to label LOOP to get the next byte. Otherwise if it is not true then the whole process of checking the alphanumeric and converting it based on the Hash table of the task starts. Finally when we find the EXIT label then we write the final result from register r5 to the memory location of r1 which is the second argument of the function returning the value.
The main problem was the missing indication of the Keil tool as there were problems in the compile that required compiler version 5.The other issue was the white spaces where we were not given any help but had to find it ourselves as the tool did not clarify where there was a problem in the gaps.
Another problem was in the first tests where the debugger did not and had the error "Error: Could not load file 'D:\DocData\school\eight\micro\keil\new1\Objects\ex1.axf'. Debugger borted !"
The testing was done by giving values for each control mentioned above to verify its correct operation. We then tested the example given in the lab pdf i.e. value = "σAr, PE 2!" and we got the result of 66 as it should be.