C CODE Generation - Why does my code not terminate?

HF
2021 2 11
1 个回答
更新时间：2021 3 8
6 次查看（30 天）

0 个投票

Hello,
I generated some C code with the Matlab coder. I wrote a main script to run the functions I generated. I dont get any error messages while running my code. However the code wont terminate. I know this question is possibly very diffcult to ask, but did someone face similar difficulties and what are possible reasons that a generated code wont terminate while I did not had any problems with the orignal code in MATLAB? I generated more then 30 functions. I dont think, its a good idea to attach all of them here.
This is my main script:
void main(void)
{
static double x_data[1024];
double Results[18];
double dv[6];
double maxP_tmp;
int Fs;
double tol;
int x_size[1];
int i;
int j;
/* Initialize function 'EV_Code' input arguments. */
/* Initialize function input argument 'x'. */
for (i = 0; i <1024; i++) {
x_data[i] = 1;
}
x_data[3] = 1.2;
x_data[230] = 1.3;
x_data[443] = 1.7;
x_data[354] = 1.9;
x_data[146] = 1.6;
x_size[0]= 2;
maxP_tmp = 1.1;
dv[0] = 114.5;
dv[1] = 90.81;
dv[2] = 73.72;
dv[3] = 87.76;
dv[4] = 72.78;
dv[5] = 59.65;
Fs = 1000;
tol = 0.04;
EV_Code_X(x_data, x_size, maxP_tmp, Fs, dv, tol, Results);
EV_Code_X_terminate();
// return 0;
}

8 个评论
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Geoff Hayes
Geoff Hayes 2021-2-11
编辑：Walter Roberson 2021-3-8
HF - what do the functions EV_Code_X and EV_Code_X_terminate do? Is it in these functions that you are getting "stuck"?
HF
HF 2021-2-11
EV_Code_X is the function where the computation is done.
EV_Code_X_terminate looks like this:
void EV_Code_X_terminate(void)
{
/* (no terminate code required) */
}
HF
HF 2021-2-11
My program is getting in EV_Code_X stuck.
HF
HF 2021-2-11
My main script is called main.c .Do I need to save my main script as a .cpp file?
HF
HF 2021-2-11
My code is getting stucked in this particular line
for (i = 0; i <= coffset; i++) {
Xm0_data[i] = Xm0->data[(aoffset + 30 * i) - 1];
}
Geoff Hayes
Geoff Hayes 2021-2-11
编辑：Walter Roberson 2021-3-8
HF- what do you mean by "getting stuck" in the above line? Note that your code has a while loop
while (exitg1 == 0);
though it does appear that your code does have a maximum number of iterations of 100.
HF
HF 2021-2-11
I paused the simulation after 1 000 000 and 2 000 000 instructions. The program was still executing this particular line.
That is the reason why I am quite confused because I had implemented a condition of maximum number of iteration for the while loop.
Rik
Rik 2021-3-8
I have restored the original question body from the Google cache. Don't remove it again.

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回答（1 个）

Geoff Hayes
Geoff Hayes 2021-2-12
编辑：Walter Roberson 2021-3-8

0 个投票

HF - the code that you mention
for (i = 0; i <= coffset; i++) {
Xm0_data[i] = Xm0->data[(aoffset + 30 * i) - 1];
}
occurs before the while loop so the while isn't the problem. In this case, what is coffset initialized to? Is it a valid or reasonable integer? It seems to get initialized via Xm0 but all I see for this variable is that it is defined as
emxArray_real_T *Xm0;
and later initialized as
emxInit_real_T(&Xm0, 2);
but is that sufficient for
i = Xm0->size[0] * Xm0->size[1];
Xm0->size[0] = 30;
Xm0->size[1] = x_size[0] + 29;
emxEnsureCapacity_real_T(Xm0, i);
? Since i is being initialized before the size, I wonder if that is somehow corrupting the object.

30 个评论
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Walter Roberson
Walter Roberson 2021-2-12
If the code for emxInit_real_T is the same as at https://www.mathworks.com/matlabcentral/answers/399098-i-am-getting-an-exception-running-c-code-generated-by-matlab-coder then the ->size would have been initialized as 0 and so i would have been initialized as 0.
HF
HF 2021-2-12
Its the same code. "i" has to be initialized before right? Otherwise I wont be able to assign a value.
void emxInit_real_T(emxArray_real_T **pEmxArray, int numDimensions)
{
emxArray_real_T *emxArray;
int i;
*pEmxArray = (emxArray_real_T *)malloc(sizeof(emxArray_real_T));
emxArray = *pEmxArray;
emxArray->data = (double *)NULL;
emxArray->numDimensions = numDimensions;
emxArray->size = (int *)malloc(sizeof(int) * numDimensions);
emxArray->allocatedSize = 0;
emxArray->canFreeData = true;
for (i = 0; i < numDimensions; i++) {
emxArray->size[i] = 0;
}
}
HF
HF 2021-2-12
Do you think I have to switch the order like this?
Xm0->size[0] = 30;
Xm0->size[1] = x_size[0] + 29;
i = Xm0->size[0] * Xm0->size[1];
HF
HF 2021-2-12
编辑：HF 2021-2-12
So I changed the order and know figured out, what the mistake is. Following equation is computed complitly wrong.
coffset = Xm0->size[1] - 2;
Xm0->size[1] has the value 30 and the value of coffset after the line is 4200000 which clearly does not make any sense. Do
Walter Roberson
Walter Roberson 2021-2-12
I don't see you printf a delimiter on the numbers ?
Walter Roberson
Walter Roberson 2021-2-12
In C and C++, printf() does not emit a space or newline, and %d format uses only the minimum number of characters needed to represent the value. Therefore if you printf("%d",3); printf("%d",4) then the output would be 34 and not 3 4 or 3<newline>4<newline>
Geoff Hayes
Geoff Hayes 2021-2-12
Try using \n as
printf("%d\n",x_data[3]);
for each printf so that the output is written to a new line.
HF
HF 2021-2-12
Thanks for the explanation. I still get a wrong value for x_data[3].
Geoff Hayes
Geoff Hayes 2021-2-12
Wait, try printing it out as a double or as a float since your x_data array is defined as a double. Use f or lf if supported:
printf("%f\n",x_data[3]);
HF
HF 2021-2-12
%f and %lf are both not supported
Walter Roberson
Walter Roberson 2021-2-12
fprintf('%d.%d\n', (int)floor(x_data[3]), (int)10*(x_data[3]-floor(x_data[3])));
HF
HF 2021-2-12
The symbol fprintf is unfortunately undefined.
Walter Roberson
Walter Roberson 2021-2-12
printf('%d.%d\n', (int)floor(x_data[3]), (int)10*(x_data[3]-floor(x_data[3])));
Walter Roberson
Walter Roberson 2021-2-12
What is the yellow caution sign warning you about?
Walter Roberson
Walter Roberson 2021-2-12
printf("%d.%d\n", (int)floor(x_data[3]), (int)10*(x_data[3]-floor(x_data[3])));
Walter Roberson
Walter Roberson 2021-2-12
At the moment I do not know how x_data[3]-floor(x_data[3]) came out negative when x_data[3] must have been positive or else there would have been a leading - as well.
Walter Roberson
Walter Roberson 2021-2-12
Your ->size are integer; you do not need to floor() them. And if you made int i then no need to (int) either.
Walter Roberson
Walter Roberson 2021-2-12
You assigned x_size[0] = 2. You set Xm0->size[1] to x_size[0]+29 and so that should give you 31. Why are you expecting 59? 59 would be Xm0->size[0] + 29 but you are grabbing x_size[0] not Xm0->size[0]
HF
HF 2021-2-12
Oh thats my mistake! Of course it should be 31
Walter Roberson
Walter Roberson 2021-2-12
Too much missing code before-hand to see.
Geoff Hayes
Geoff Hayes 2021-2-12
编辑：Walter Roberson 2021-3-8
HF- perhaps clarify what the problem might be or is it still the same as before. Are you still getting stuck in that loop. If so, have you written out what the upper bound on the loop is?
HF
HF 2021-2-12
编辑：HF 2021-2-12
Exactly, so I still get stucked in the loop I mentioned before.
for (i = 0; i <= coffset; i++) {
Xm0_data[i] = Xm0->data[(aoffset + 30 * i) - 1];
}
Coffset has a value of 4000000 which does not make sense. As you can see from the snapshots before, the calculation of several equations are incorrect. Now I am trying to figure out why this might be the case.
HF
HF 2021-2-12
My main approach is to display every value after an equation. As you can see in the snapshot the first results of the function EV_Code_X does not make sense.
I hope this clarifies my problem.
Geoff Hayes
Geoff Hayes 2021-2-12
Out of curiosity, where or when is memory allocated to Xm0->data? It seems that it is initialized as
emxArray->data = (double *)NULL;
(in the emxInit_real_T function) but where is the memory for it allocated? Since the code seems to read and write data to this array I wonder if the writing to this array before memory has been allocated is corrupting the other data in this object...
Geoff Hayes
Geoff Hayes 2021-2-12
编辑：Geoff Hayes 2021-2-12
Or is that what emxEnsureCapacity_real_T does, and if so, are we passing in the correct capacity i? From this code
// i = Xm0->size[0] * Xm0->size[1];
Xm0->size[0] = 30; // 30
printf("%d\n", (Xm0->size[0])); //H
Xm0->size[1] = x_size[0] + 29; // 31 (assuming x_size[0] is 2)
printf("%d\n", Xm0->size[1]); //H
i = Xm0->size[0] * Xm0->size[1]; // 30 * 31 = 930
printf("%d\n", (i));//H
emxEnsureCapacity_real_T(Xm0, i); // allocated for 930 elements
loop_ub = 30 * (x_size[0] + 29); // 30 * (2+29) = 930
for (i = 0; i < loop_ub; i++) {
Xm0->data[i] = 0.0; // this "seems" fine
}
/* y = f*x where x is padded */
for (aoffset = 0; aoffset < 30; aoffset++) {
if (aoffset + 1 == 1) {
if (1 > N) {
loop_ub = 0;
} else {
loop_ub = N; // N is 2?
}
// this occurs on first iteration of loop
for (i = 0; i < loop_ub; i++) {
Xm0->data[30 * i] = x_data[i]; // i is 0 or 30
}
// etc.
HF
HF 2021-2-12
Xm0->data is initialized in the following line (in the beginning of the function EV_X_Code):
for (i = 0; i < loop_ub; i++) {
Xm0->data[i] = 0.0;
}
HF
HF 2021-2-12
We are passing the right capacit of i. I mean everything before emxEnsureCapacity_real_T(Xm0, i); should work correclty. However printf("%d\n", (i));//H does not display the correct value.
To your comments: N = 2, i is 0 for the first loop.
Geoff Hayes
Geoff Hayes 2021-2-12
The passing of i in emxEnsureCapacity_real_T(Xm0, i) may be irrelevant because of
if (emxArray->data != NULL) {
and this should be null because it is initialized to null in the other function at
emxArray->data = (double *)NULL;
So there is no data to copy into the newly allocated array, so this parameter of i is unimportant.
But in this capacity function, look at this code
newNumel = 1;
for (i = 0; i < emxArray->numDimensions; i++) {
newNumel *= emxArray->size[i];
}
if (newNumel > emxArray->allocatedSize) {
i = emxArray->allocatedSize;
if (i < 16) {
i = 16;
}
while (i < newNumel) {
if (i > 1073741823) {
i = MAX_int32_T;
} else {
i *= 2;
}
}
newData = calloc((unsigned int)i, sizeof(double));
Given what we already know, newNumel should be 30*31=930. The emxArray->allocatedSize is 0 so i is initialized to 16. Then in the while loop we multiply i by 2 on each iteration until it is greater than newNumel and so should be initialized to 1024 which you should be able to confirm with Xm0->allocatedSize. So you should have enough memory allocated in your array. Can you confirm that this allocatedSize is 1024?
Geoff Hayes
Geoff Hayes 2021-2-14
Why do you think that the program skipped the call to emxEnsureCapacity_real_T? Check the Xm0->allocatedSize to see what this value is. (I don't have the MATLAB coder so cannot test out your files.)

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HF
HF
2021-2-11

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Walter Roberson
Walter Roberson
2021-3-8

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