ryan
Reputation: 153
Double free or corruption: C++
So I am getting a memory leak error from my code:
*** glibc detected *** ./KalmanFiltering: double free or corruption (!prev): 0x00000000015af7b0 ***
======= Backtrace: =========
/lib/x86_64-linux-gnu/libc.so.6(+0x7eb96)[0x7f0897395b96]
./KalmanFiltering[0x40654d]
/lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xed)[0x7f089733876d]
./KalmanFiltering[0x4012b9]
======= Memory map: ========
00400000-00415000 r-xp 00000000 00:15 6312794 /home/iggy/Dropbox/Documents/Research_Work/SimpleHealth/KalmanFilter/KalmanFilter_C++/cmpfit-1.2/KalmanFiltering
00614000-00615000 r--p 00014000 00:15 6312794 /home/iggy/Dropbox/Documents/Research_Work/SimpleHealth/KalmanFilter/KalmanFilter_C++/cmpfit-1.2/KalmanFiltering
00615000-00616000 rw-p 00015000 00:15 6312794 /home/iggy/Dropbox/Documents/Research_Work/SimpleHealth/KalmanFilter/KalmanFilter_C++/cmpfit-1.2/KalmanFiltering
015ae000-01641000 rw-p 00000000 00:00 0 [heap]
7f0897317000-7f08974cc000 r-xp 00000000 08:01 421630 /lib/x86_64-linux-gnu/libc-2.15.so
7f08974cc000-7f08976cc000 ---p 001b5000 08:01 421630 /lib/x86_64-linux-gnu/libc-2.15.so
7f08976cc000-7f08976d0000 r--p 001b5000 08:01 421630 /lib/x86_64-linux-gnu/libc-2.15.so
7f08976d0000-7f08976d2000 rw-p 001b9000 08:01 421630 /lib/x86_64-linux-gnu/libc-2.15.so
7f08976d2000-7f08976d7000 rw-p 00000000 00:00 0
7f08976d7000-7f08976ec000 r-xp 00000000 08:01 395568 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f08976ec000-7f08978eb000 ---p 00015000 08:01 395568 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f08978eb000-7f08978ec000 r--p 00014000 08:01 395568 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f08978ec000-7f08978ed000 rw-p 00015000 08:01 395568 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f08978ed000-7f08979e8000 r-xp 00000000 08:01 422139 /lib/x86_64-linux-gnu/libm-2.15.so
7f08979e8000-7f0897be7000 ---p 000fb000 08:01 422139 /lib/x86_64-linux-gnu/libm-2.15.so
7f0897be7000-7f0897be8000 r--p 000fa000 08:01 422139 /lib/x86_64-linux-gnu/libm-2.15.so
7f0897be8000-7f0897be9000 rw-p 000fb000 08:01 422139 /lib/x86_64-linux-gnu/libm-2.15.so
7f0897be9000-7f0897ccb000 r-xp 00000000 08:01 531352 /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.16
7f0897ccb000-7f0897eca000 ---p 000e2000 08:01 531352 /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.16
7f0897eca000-7f0897ed2000 r--p 000e1000 08:01 531352 /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.16
7f0897ed2000-7f0897ed4000 rw-p 000e9000 08:01 531352 /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.16
7f0897ed4000-7f0897ee9000 rw-p 00000000 00:00 0
7f0897ee9000-7f0897f0b000 r-xp 00000000 08:01 422388 /lib/x86_64-linux-gnu/ld-2.15.so
7f08980a3000-7f08980ea000 rw-p 00000000 00:00 0
7f0898107000-7f089810b000 rw-p 00000000 00:00 0
7f089810b000-7f089810c000 r--p 00022000 08:01 422388 /lib/x86_64-linux-gnu/ld-2.15.so
7f089810c000-7f089810e000 rw-p 00023000 08:01 422388 /lib/x86_64-linux-gnu/ld-2.15.so
7fffc58ae000-7fffc58cf000 rw-p 00000000 00:00 0 [stack]
7fffc59af000-7fffc59b0000 r-xp 00000000 00:00 0 [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0 [vsyscall]
Aborted (core dumped)
Running where in gdb I get:
#0 0x00007ffff723e425 in __GI_raise (sig=<optimized out>) at ../nptl/sysdeps/unix/sysv/linux/raise.c:64
#1 0x00007ffff7241b8b in __GI_abort () at abort.c:91
#2 0x00007ffff727c39e in __libc_message (do_abort=2, fmt=0x7ffff7386748 "*** glibc detected *** %s: %s: 0x%s ***\n") at ../sysdeps/unix/sysv/linux/libc_fatal.c:201
#3 0x00007ffff7286b96 in malloc_printerr (action=3, str=0x7ffff7386858 "double free or corruption (!prev)", ptr=<optimized out>) at malloc.c:5039
#4 0x000000000040654d in main (argc=6, argv=0x7fffffffe0c8) at KalmanFiltering.cpp:828
where 828 is the return line in my main() function.
The code in question is:
int main(){
...
EKSmoothParams *EKParams = new EKSmoothParams;
prepareSmoother(optPar, ECGsd, peaks, phase, x, fs, EKParams);
delete EKParams;
return 0;
}
void prepareSmoother(vector<double> optPar, vector<double> ECGsd, vector<double> peaks, vector<double> phase, vector<double> x, double fs, EKSmoothParams *params){
const int N = PEAK_NUM; // number of Gaussian kernels
vector<int> JJ;
JJ.reserve(peaks.size());
for(int i = 0; i < peaks.size(); i++){
if(peaks.at(i) != 0)
JJ.push_back(i);
}
vector<double> fm; // heart-rate
fm.reserve(JJ.size()-1);
for(int i = 0; i < JJ.size()-1; i++){
fm.push_back(fs/(JJ.at(i+1)-JJ.at(i)));
}
vector<double> twoPiFm = fm;
for(int i = 0; i < fm.size(); i++)
twoPiFm[i] = 2*PI*fm.at(i);
double w = calculateMean(twoPiFm); // average heart-rate in rads.
double wsd = sd2(twoPiFm); // heart-rate standard deviation in rads.
params->X0[0][0] = 1.0;
params->X0[0][1] = -PI;
params->X0[1][0] = 1.0;
params->X0[1][1] = 0.0;
params->P0[0][0] = pow(2*PI,2);
params->P0[0][1] = 0.0;
params->P0[1][0] = 2.0;
params->P0[1][1] = 10*pow(findAbsMax(x),2.0);
vector<double> diagonal(3*N+2, 0.0);
for(int i = 0; i < N; i++)
diagonal[i] = pow(0.1*optPar.at(i),2.0);
for(int i = N; i < 3*N; i++)
diagonal[i] = pow(0.5,2.0);
diagonal[3*N] = pow(wsd,2.0);
vector<double>::const_iterator first = ECGsd.begin();
vector<double>::const_iterator last = ECGsd.begin() + round(ECGsd.size()/10.0);
vector<double> ECGsdPartial(first, last);
displayVector(ECGsd);
diagonal[3*N+1] = pow(0.05*calculateMean(ECGsdPartial), 2.0);
for(int i = 0; i < diagonal.size(); i++)
params->Q[i][i] = diagonal[i];
params->R[0][0] = pow(w/fs,2)/12.0;;
params->R[0][1] = 0.0;
params->R[1][0] = 0.0;
params->R[1][1] = pow(calculateMean(ECGsdPartial), 2.0);
for(int i = 0; i < optPar.size(); i++){
params->wMean[i] = optPar.at(i);
params->inits[i] = optPar.at(i);
}
params->wMean[N*3] = w;
params->wMean[N*3+1] = 0;
params->inits[N*3] = w;
params->inits[N*3+1] = fs;
params->vMean[0] = 0.0;
params->vMean[1] = 0.0;
params->inovWlen = round(0.5*fs+0.5);
params->tau = 0;
params->gamma = 1;
params->rAdaptWlen = round(fs/2.0 + 0.5);
params->flag = 1;
for(int i = 0; i < 2; i++){
for(int j = 0; j < 2; j++){
cout << params->R[i][j] << " ";
}
}
}
And the struct is statically allocated:
struct EKSmoothParams {
int tau;
int gamma;
int flag;
int inovWlen;
int rAdaptWlen;
double wMean[3*PEAK_NUM+2];
double vMean[2];
double inits[3*PEAK_NUM+2];
double X0[2][2];
double P0[2][2];
double Q[3*PEAK_NUM+2][3*PEAK_NUM+2];
double R[2][2];
};
where:
#define MEAN_PEAK_NUM_HIGH 3
#define MEAN_PEAK_NUM_LOW 3
#define PEAK_NUM MEAN_PEAK_NUM_HIGH+MEAN_PEAK_NUM_LOW
Any help would be greatly appreciated. I thought that if a struct was statically allocated then only calling
delete struct_name;
would delete the reference. Thanks for the help!
UPDATE:
So I ran valgrind and it told me to remove the delete EKParms. I removed the delete EKParms line and ran it again as:
valgrind --leak-check=full ./KalmanFiltering x.txt pphase.txt phase.txt opt.txt peaks.txt
and the output i got was:
==32755== Memcheck, a memory error detector
==32755== Copyright (C) 2002-2011, and GNU GPL'd, by Julian Seward et al.
==32755== Using Valgrind-3.7.0 and LibVEX; rerun with -h for copyright info
==32755== Command: ./KalmanFiltering x.txt pphase.txt phase.txt opt.txt peaks.txt
==32755==
0.117777 0.102793 0.0911142 0.107277 0.109 0.126729 0.115012 0.109627 0.131102 0.100467 0.0822639 0.122442 0.0908527 0.116644 0.108093 0.104796 0.12665 0.102979 0.0999146 0.119981 0.107912 0.122379 0.113098 0.0889197 0.106954 0.101472 0.125473 0.107778 0.13228 0.10528 0.11511 0.107965 0.0961817 0.125068 0.12075 0.110846 0.120621 0.132226 0.106999 0.114672 0.0997654 0.104048 0.117857 0.105461 0.127318 0.11103 0.134415 0.12594 0.126633 0.116603 0.109422 0.117 0.130797 0.112808 0.113414 0.0951991 0.112291 0.109693 0.118444 0.104215 0.124635 0.0993083 0.122034 0.122363 0.12139 0.0969221 0.108173 0.109436 0.115881 0.118631 0.0968963 0.104841 0.118923 0.10789 0.108117 0.119053 0.115187 0.119369 0.089593 0.0893818 0.127805 0.109007 0.108001 0.128517 0.105524 0.117847 0.127699 0.101618 0.113646 0.112389 0.114674 0.108706 0.117413 0.119509 0.110195 0.116943 0.132244 0.108374 0.117175 0.114302 0.113753 0.127603 0.104102 0.112583 0.110015 0.102419 0.122587 0.104333 0.122883 0.129287 0.129104 0.10733 0.11312 0.125945 0.119181 0.128817 0.129468 0.114589 0.146289 0.135648 0.118936 0.146207 0.160105 0.167322 0.16074 0.140913 0.153806 0.178517 0.215381 0.251905 0.163988 0.137749 0.107549 0.121755 0.121141 0.0867208 0.103768 0.130058 0.142986 0.115026 0.12086 0.12443 0.122726 0.110762 0.125137 0.126337 0.0953488 0.10774 0.112677 0.116888 0.115948 0.104844 0.114403 0.121069 0.110119 0.0980817 0.109335 0.104094 0.10667 0.118813 0.123157 0.11163 0.105456 0.103909 0.112385 0.126633 0.123956 0.108601 0.113358 0.0971531 0.123609 0.116769 0.130958 0.103691 0.114814 0.116871 0.12273 0.116116 0.118833 0.11895 0.100572 0.128861 0.110058 0.121104 0.122787 0.122287 0.114645 0.12352 0.122679 0.121228 0.116913 0.128488 0.111704 0.102892 0.119502 0.113897 0.144082 0.132502 0.115685 0.145348 0.137543 0.12479 0.132752 0.137675 0.144116 0.127518 0.146219 0.152045 0.123085 0.152635 0.153129 0.159488 0.139282 0.150634 0.119596 0.1195 0.127458 0.109524 0.106355 0.116666 0.114375 0.104727 0.0978894 0.0941401 0.11789 0.11224 0.110342 0.106331 0.104715 0.0991576 0.116447 0.0908483 0.11542 0.105876 0.0955746 0.120995 0.125514 0.130953 0.12472 0.118668 0.118989 0.106662 0.117213 0.111635 0.106181 0.11708 0.101769 0.10301 0.112952 0.104064
3.31532e-06 0 0 0.0118791
==32755==
==32755== HEAP SUMMARY:
==32755== in use at exit: 0 bytes in 0 blocks
==32755== total heap usage: 61,624 allocs, 61,624 frees, 6,091,874 bytes allocated
==32755==
==32755== All heap blocks were freed -- no leaks are possible
==32755==
==32755== For counts of detected and suppressed errors, rerun with: -v
==32755== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)
However, when I run the program normally I still get the double free or corruption error as before.
Upvotes: 0
Views: 2520
Answers (1)
hyde
Reputation: 62777
Problem is probably memory corruption. You have buggy #define PEAK_NUM. Needs parenthesis:
#define PEAK_NUM (MEAN_PEAK_NUM_HIGH+MEAN_PEAK_NUM_LOW)
Without those extra parenthesis, you for example have this:
double Q[3*MEAN_PEAK_NUM_HIGH+MEAN_PEAK_NUM_LOW+2][3*MEAN_PEAK_NUM_HIGH+MEAN_PEAK_NUM_LOW+2];
And that obviously is much less than you expect, so your function probably corrupts memory and anything can happen after that.
Start by fixing that! Also, this demonstrates why macros are evil. You have to be very careful when using them, and still they occasionally bit even an experienced programmer.
Even though this is out of the scope of the question, here's a way to more safely define the constants:
enum {
MEAN_PEAK_NUM_HIGH = 3,
MEAN_PEAK_NUM_LOW = 3,
PEAK_NUM = MEAN_PEAK_NUM_HIGH + MEAN_PEAK_NUM_LOW
};
Upvotes: 4