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OK, I know what you mean 
For me, it is easier to use WPF because I am embedded system engineer.
I am from embedded world and I do not know Windows internals.
That is why I am asking a lot of (stupid) questions. After reading several articles I supposed that redirect output and develop GUI app should be easier
Now I know that my strange console app does not use standard printf function 
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The obvious question is can you not simply replace the program communicating with a proper windows code or is it a private format they won't disclose. Just saying it would be faster, cleaner and less code than what you are doing now
In vino veritas
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Yes, it is a private closed format.
Another way to communicate with the console app is passing command line arguments:
MyApp.exe -Iinput.txt -Ooutput.txt
where in the input.txt file in each line is a key command (e - exits the program, other keys do some actions):
l
f
m
c
o
e
in the output.txt file, the program saves its output. Even something like this works:
MyApp.exe -OCON - prints output to console twice.
But problem is the same. If I monitor the output file, the content appears after app quits (e command in input file)
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This smells to me of buffering. I'm guessing your app doesn't call flush() or equivalent on its output file, so the output is buffered until the explicit or implicit close() call at termination.
If you can make your app output lots of data, you will probably see it appear in chunks, which will be whatever buffer size is used.
The reason it works on the console is that console drivers do not buffer.
I have no idea if you can suppress buffering in your redirection, but if you could, my 2c says that would solve your problem.
Good luck!
Peter
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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Yes standard console output is always buffered.
You can actually easily remove it
setvbuf(stdout, NULL, _IONBF, 0);
setvbuf(stderr, NULL, _IONBF, 0);
The problem is it will slow the console program down like a dog as each character invokes a full call.
I sometimes do it when I have a embedded target with UART debugger like on the Raspberry Pi.
In vino veritas
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my code is:
#include<stdio.h>
#include<conio.h>
struct point
{
int x,y;
void getdis()
{
printf("%d",x+y);
}
};
int main()
{
struct point p;
int a=5,b=10;
p.getdis(a,b);
getch();
return 0;
}
problem is:8 2 F:\software\Dev-Cpp\bin\point.c [Error] expected ':', ',', ';', '}' or '__attribute__' before '{' token
I cannot understand what to write before the getdis function to avoid this error
suman chandra modak
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void getdis()
...
p.getdis(a,b);
These two do not match. Hint: they should.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
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methods are a C++ feature you can only mimic in C, suing function pointers and the like. For instance
#include<stdio.h>
struct point
{
int x,y;
void (*pfun)(struct point *pp);
};
void getdis(struct point *pp)
{
printf("%d\n", (pp->x + pp->y));
}
int main()
{
struct point p;
p.x = 5; p.y = 10;
p.pfun = getdis;
p.pfun(&p);
getchar();
return 0;
}
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i am trying to write a code to solve Laplace equation using Successive
Over relaxation(SOR)on a NXN matrix
boundary condition is X^2-y^2
i dont know how to set this boundary condition on the Mesh created by me
here is the code
import numpy as np
from pylab import *
import sys
import copy
import matplotlib.pyplot as plt
import matplotlib.ticker as ticker
L =[[]for i in range(2)]
N=int(input("Enter a number N for N x N grid: "))
h=1/N
def clear_list1():
L[1].clear()
L[0].clear()
iterat=0
error=0.75
clear_list1()
xx=np.arange(0,N,h)
yy=np.arange(0,N,h)
x,y = np.meshgrid(xx,yy)
phi= copy.deepcopy((x**2-y**2))
phi[1:-1,1:-1]=0
phi_new = copy.deepcopy(phi)
resid=np.zeros((N,N))
while True:
if iterat<21:
for w in np.linspace(1,2,11):
for i in range(1,N-1):
for j in range(1,N-1):
phi1=(phi[i+1,j]+phi[i-1,j]+phi[i,j+1]+phi[i,j-1])/4
phi[i,j]=w*phi1+(1-w)*phi1
Sumresidual=0
for i in range(1,N-1):
for j in range(1,N-1):
resid[i,j]= -4*phi[i,j]+phi[i+1,j]+phi[i-1,j]+phi[i,j+1]+phi[i,j-1]
Sumresidual=Sumresidual+(resid[i,j])**2
else:
break
iterat=iterat+1
p=np.sqrt(Sumresidual)
L[0].append(p)
L[1].append(w)
ax = plt.axes()
ax.xaxis.set_major_locator(ticker.MaxNLocator(5))
plt.plot(L[0],L[1])
plt.yscale('log')
plt.xlabel('w')
plt.ylabel('|| residual ||')
plt.title('effect of w on residual')
plt.show()
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And what does your code have to do with the C / C++ / MFC Forum?
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hi
I was getting storage overlays in my local variable when I noticed that the SP register was 0X4652E25F0 and my BP registers was 0X4C652E2A60 does that mean that my stack size is 144 bytes
I have a table in the function itself that is 256 bytes to translate ASCII to EBCDIC
beginthread stack size is 0
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Quote: In addition, the stack_size argument can be 0, in which case the operating system uses the same value as the stack that's specified for the main thread.
Don't try to guess what the situation may be just by looking at registers. In most code BP is set as a base register from an offset of SP, but that does not imply any particular stack size.
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I guess it could be set /F compiler option thanks
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ForNow wrote: I guess it could be ...
... anything. Guessing is not generally a good strategy for fixing problems. You need to be sure what the problem is first. Maybe if you show the code where the problem occurs, and a few more details, we could make some useful suggestions.
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if stepped thru the code in Assembler after the function prologue by subtracting BP - SP = stack size right ?
I added code to a source file in Hercules cpu.c thing is its a makefile build which very confusing I was wondering if there isn't -F or /F flag for the stack size I guess it would default to some number my local variables and parameters would make the stack a little over 300 bytes, which is not too big
I am trying to figure to determine where in the makefile is the compile for CPU.C
Thanks
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I think we are talking at cross-purposes. What you are referring to is a function's frame size, which is a portion of the thread's stack. The frame will be created large enough for all the locally declared variables in that function, so you need to look at the source code to see why it is not large enough, or why it is being overwritten.
The makefile may be using an inference rule for cpu.c, so you cannot always find a specific line which references it. But you should be able to find a reference to it somewhere in the Makefile (or any of its included subfiles). However, as I mentioned, you still need to diagnose the actual problem, and changing the stack size for the final application is unlikely to make any difference.
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I understand finally sorry I’m slow thanks so much
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Dear Respected friends.
good morning. i hope you are fine.
I need your help. I have a problem with smoothing of output. I have this code ..
it exports stair stepping.
Can you help me.?
int Imax = 10;
for ( int I=0; I<imax; i++="" )="" {
="" for="" (int="" iy="0;" iy<ny;="" iy++)="" {
="" ix="0;" ix<nx;="" ix++)="" float="" v1="v[(iy*nx+ix)*nz];
" int="" k1="0;
" k0="0;
" jstart="0;
" while="" (k1="" <="" nz)="" k;
="" (k="k1;" k<nz;="" k++)="" v2="v[(iy*nx+ix)*nz" +="" k];
="" if="" (="" 100.="" *="" fabs(v2="" -v1)="">= sharpeningThreshold_percentage/2. && jstart == 0 ) {
k0 = k;
jstart = 1;
}
if ( 100. * fabs(v2 -v1) / v1 >= sharpeningThreshold_percentage ) {
v1 = v2;
k1 = k;
jstart = 0;
for ( int kk=k0; kk<=k; kk++ )
v[(iy*nx+ix)*nz + kk] = v1;
k0 = k1;
break;
}
v[(iy*nx+ix)*nz + k] = v1;
}
if ( k >= nz-1 )
break;
}
}
}
if ( I < Imax-1 )
smooth(10, 1, v, nx, ny, nz);
}
}
}
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You did not state what the problem is. What are you expecting from your code? What do you get instead that makes you say it is problematic?
"I'm neither for nor against, on the contrary." John Middle
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Dear Friend.
The code works. But the output comes as stair steps. But, this is because of this sharpening .. and I want to alter . if you can give idea how to make the smoothing.
v2 is the velocity of second layer. and v1 is velocity of first layer.
maximum iterations is 10 with the percentage of threshold ..
problem is that it gives sharp stair stepping. while I want to make like smooth signal in output.
If you can guide.
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Hello,
I got problem trying converting type when the var is declared void in function.
How do you do these things !? for example casting short to int from void var (declared short at start)
And can we change a variable type without having to create new var !?
4 days im on, and i dont get solution without rewriting 50 "identical" functions just with changed type...
It's for that im looking to the void type.
Please help 
#include <stdio.h>
void calc(void *var) {
int tmp = *(int*)var;
printf("tmp: %i\n", tmp); *(int*)var += 30003;
}
void complexeShortToInt() {
short int vInt2 = 30002;
calc(&vInt2);
printf("complexeShortToInt: %i\n", vInt2);}
int main() {
complexeShortToInt(); return 0;
}
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Do not use void when you know the type that is to be operated on, it makes no sense. And your calc function would be better to return the calculated value, something like:
int calc(int var) {
int tmp = var + 30003;
return tmp;
}
I am not sure what the complexeShortToInt function, or the actual program, is supposed to do.
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Surely it is not to put these sh*tty functions in my program, but it is to understand how i can handle this situation (so i reduced the code at max)
And the real situation of calc is to substitute the standard + - * / operations with "handle" of the overflow.
So for example a declared short int that excess the 32767 limit.
I know the type at start, but maybe it will produce an overflow later, and maybe with difficulty to detect.
So how we handle this situation !?
Here another more complete code, that can represent a var++;
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
void calc(char *type, void *var1, char calc, void *var2) {
int tmp1 = *(int*)var1;int tmp2 = *(int*)var2;
printf("tmp1: %i\n", tmp1);
printf("tmp2: %i\n", tmp2);
int max, min;
unsigned int umax, umin;
short int overflow = 0;
if(strncmp(type, "short int", 9)==0) {
max = SHRT_MAX;
min = SHRT_MIN;
}
if(calc == '+') {
}
}
int main() {
short int vInt2 = 32767; short int vInt3 = 1;
calc("short int", &vInt2, '+', &vInt3);
printf("bad convert to int: %i\n", vInt2);
return 0;
}
Edit: So if i dont use void, it seems i have to write lot of calc "identical" function for each type....
modified 10-Feb-18 14:26pm.
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You misunderstand the problem I'm afraid. Using a void pointer makes no difference, as your pointers will still point to different types. For example:
void calc(void* pvoid)
{
int value = *(int*)pvoid:
}
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