bools in python

[vajratkarviraj]

question from a newbie:
can somebody help me in converting this c++ code to python...

bool flag;
do{        
flag=0;      
for(int i=0;i<n;i++){
        double s1=xold[i];
        double s2=xnew[i];  
        e[i]= s1-s2;
        if (e[i]<0) {e[i]=0-e[i];}
        if(e[i]>prec){flag=1;}
 
for(int i=0;i<n;i++){
        xold[i]=xnew[i];
        }
 
}
}while (flag);
 

basically the bool thing is a prob... also if i hav created a function(say solve(parameters)) within a function(say display(parameters)).... then the python code goes sumwat like this:

 
def display(parameter1, parameter2):
 
{code of display}
 
solve(parameter3, parameter4):
 
{code of solve and it gives me output its output in the form of a number}
 
#now i want to return control to display
 
{remaining code of display continues...}
 

[bartonc]

question from a newbie:
can somebody help me in converting this c++ code to python...

bool flag;
do{        
flag=0;      
for(int i=0;i<n;i++){
        double s1=xold[i];
        double s2=xnew[i];  
        e[i]= s1-s2;
        if (e[i]<0) {e[i]=0-e[i];}
        if(e[i]>prec){flag=1;}
 
for(int i=0;i<n;i++){
        xold[i]=xnew[i];
        }
 
}
}while (flag);
 

<snip>

I'm not so hot with C, but it looks like the while will never execute due to flag=0, but (or maybe it executes once because while() is evaluated last) - I'll show the latter:

n = 4 # counter for for()
flag = 0
while True:   # Note that python is indentation based
    for i in range(n):  # use xrange() for huge numbers
        s1 = xold[i]
        s2 = xnew[i]
        e[i] = s1 - s2
        if e[i] < 0:   # colon starts a block, next de-dent ends it
            e[i] = 0 - e[i]
        if e[i] > prec:
            flag=1
       # comments can go on any indent level - etc...
    if not flag:
        break

That is, if I'm reading the C part correctly. Close, anyway

[bartonc]

I'm not so hot with C, but it looks like the while will never execute due to flag=0, but (or maybe it executes once because while() is evaluated last) - I'll show the latter:

n = 4 # counter for for()
flag = 0
while True:   # Note that python is indentation based
    for i in range(n):  # use xrange() for huge numbers
        s1 = xold[i]
        s2 = xnew[i]
        e[i] = s1 - s2
        if e[i] < 0:   # colon starts a block, next de-dent ends it
            e[i] = 0 - e[i]
        if e[i] > prec:
            flag=1
       # comments can go on any indent level - etc...
    if not flag:
        break

That is, if I'm reading the C part correctly. Close, anyway

Anything that has a non-zero value will evaluate as True.
bool is a subclass of int so you can do many things with python bools that other languages wouldn't dream of doing (including list indexing):

timeToStop = True
print ['still going', 'stopping'][timeToStop]

Kind of nifty, huh?

[Smygis]

question from a newbie:
can somebody help me in converting this c++ code to python...

bool flag;
do{        
flag=0;      
for(int i=0;i<n;i++){
        double s1=xold[i];
        double s2=xnew[i];  
        e[i]= s1-s2;
        if (e[i]<0) {e[i]=0-e[i];}
        if(e[i]>prec){flag=1;}
 
for(int i=0;i<n;i++){
        xold[i]=xnew[i];
        }
 
}
}while (flag);
 

basically the bool thing is a prob... also if i hav created a function(say solve(parameters)) within a function(say display(parameters)).... then the python code goes sumwat like this:

 
def display(parameter1, parameter2):
 
{code of display}
 
solve(parameter3, parameter4):
 
{code of solve and it gives me output its output in the form of a number}
 
#now i want to return control to display
 
{remaining code of display continues...}
 

First of, Only becouse C++ lets you make your code 100% unreadable does not mean you shuld do it.

bool flag;
do{        
    flag = 0;
    for(int i = 0; i < n; i++)
    {
        double s1 = xold[i];
        double s2 = xnew[i];
 
        e[i] = s1 - s2;
 
        if (e[i] < 0)
            e[i] = 0 - e[i];
 
        if(e[i] > prec)
            flag = 1;
 
        for(int i = 0; i < n; i++)
            xold[i] = xnew[i];
    }
}while(flag);

And as for your second Q,

def solver(arg1, arg2):
    return arg1 ** arg2 # Ofcource you will replace this with whatever you want it to do
 
def display(no1, no2):
    # Code
    solvednum = solver(no2, no1)
    # more code
 

Simply create multiple functions.