[WMCoolmon]

Originally posted by kasperl
Bobb: What about doing what the TI calculators do and use a different minus sign for the negative and the substraction? Use an underscore or something for one of them.

It'd be a real pain to do that since you'd have to open up charmap every time you wanted to use a negative.

Although I suppose tilde or something could be subbed in...it'd make the code less readable though.

[kasperl]

I know it isn't fun for readability, but it'd save a ****load of coding, and quite a few ambiguities.

[Anaz]

or do something like using tilde or another character as your "negative" minus sign rather than subtraction operation.

[Taristin]

screw readability, once you get it working once, don't touch it again

[Bobboau]

I think I'm going to have any time it finds a '-' without an operand on either side it'll replace it with '-1*', sence -# is working

[StratComm]

what about (a)-(b)?

[Bobboau]

this change fixes it I think.

it actualy replaces any '-' that does not have a valid charicter before it with '(-1*' and finds the end of the block it's being multiplyed with and adds a ')'
so -x is replaced with (-1*x)
-(14*5+3) is replaced with (-1*(14*5+3))

Code: [Select]

void fix_implied_mult(string &str){
for(int i = 0; i< str.length(); i++){
if(i && str[i] == '('){
if(!get_value(str[i-1]))
str.insert(i,"*");
}
if(str[i] == '-'){
if((!i || (!is_value(str[i-1]) && str[i-1] != ')') )){
str.insert(i,"(");
int p = 0;
for(int k = i+2; get_value(str[k]) == -1 || p; k++){
if(str[k] == '(')p++;
if(str[k] == ')')p--;
}
str.insert(k,")");
str.insert(i+2,"1*");
i+=3;
}
}
}
}


[Bobboau]

test build!

[Scuddie]

Code: [Select]

enter an expression or type exit to do so
-(9+11)

This application has requested the Runtime to terminate it in an unusual way.
Please contact the application's support team for more information.

[Bobboau]

Code: [Select]

void fix_implied_mult(string &str){
for(int i = 0; i< str.length(); i++){
if(i && str[i] == '('){
if(!get_value(str[i-1]))
str.insert(i,"*");
}
if(str[i] == '-'){
if((!i || (!is_value(str[i-1]) && str[i-1] != ')') )){
str.insert(i,"(");
int p = 0;
for(int k = i+2; (get_value(str[k]) == -1 || p) && str[k]; k++){
if(str[k] == '(')p++;
if(str[k] == ')')p--;
}
str.insert(k,")");
str.insert(i+2,"1*");
i+=3;
}
}
}
}


ok, try again

[Drew]

if i put in -1 by itself, the thing crashes

[Drew]

-x by itself crashes in general

[Scuddie]

Code: [Select]

enter an expression or type exit to do so
-(9+11)

-(9+11) = -20
if this is not corect, please tell me and provide the expression you entered

enter another expression or type exit to do so:
2^-(1)

2^-(1) = 0.5
if this is not corect, please tell me and provide the expression you enteredI think you got that part down, bob .

[Flipside]

-((2/4)*3+(2^2)+6)-3 = -14.5

This seems wrong....

=   -((0.5)*3+(4)+6)-3
= -(1.5)+(10)-3
= -8.5

Though my maths brain cells haven't been used in years, so it could be down to my knowledge of BODMAS It was the only calculation that seemed a bit off.

EDIT : This was prefix 5, I'll try 6 now Still -14.5. Someone check my maths, I'm still not certain

[Bobboau]

someone pointed out something screwed up, new version better implyed multiplication suport now.

[Scuddie]

Sorry flip, your math is wrong .  You got the first line right, but you will notice how you made your mistake.

= -((0.5*3)+4+6)-3
= -((1.5)+10)-3
= -(11.5)-3
= -14.5

There was your problem.  As my old math teacher said, "the most important part of an equation is the attention to detail".

[Flipside]

Ahhh... yes, the -3 is outside the brackets and so applies after the equation has been inversed I'd need another set of brackets to get the result I got

Ah well, told ya it'd been a while

[Bobboau]

because I'm on a computer were I can't save, I'm uploading my code

[WMCoolmon]

Compiled version (EXE):

[Bobboau]

I'm going to assume that the silence means no one has found any more errors...