Why does compiler throw "terminate called recursively"

user_15794882 0
Viewed 143

I'm trying to experiment passing 2D arrays to a function using C++. As part of that, when I allocated the space to a 3*2 matrix, and during execution, I'm not able to figure out as to why the compiler throws "terminate called recusively".

Any help would be appreciated.

Edit: This works perfectly fine in Online compiler. (My system used a gcc compiler -> MingW on Windows)

void printMat(int **arr, int m, int n)
{
    for(int i=0; i<m; i++)
        for(int j=0; j<n; j++)
            cout<<arr[i][j]<<" ";
}

int main()
{
    int m=3, n=2;

    int **arr;
    arr = new int *[m];

    for(int i=0; i<m; i++)
    {
        arr[i] = new int[n];
        for(int j=0; j<m; j++)
        {
            arr[i][j]=i;
            cout<<arr[i][j]<<" ";
        }
    }
    cout<<"Reached"<<endl;
    printMat(arr, m, n);
    return 0;
}
c++14 pointers c++
1 Answers

I strongly recommend using sanitizers in all debug builds to track these bugs, see live example that quickly tells you what is wrong.

=================================================================
==1==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x602000000018 at pc 0x000000401752 bp 0x7ffeb62d0c60 sp 0x7ffeb62d0c58
WRITE of size 4 at 0x602000000018 thread T0
    #0 0x401751 in main /app/example.cpp:22
    #1 0x7f48243620b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
    #2 0x4011dd in _start (/app/output.s+0x4011dd)

0x602000000018 is located 0 bytes to the right of 8-byte region [0x602000000010,0x602000000018)
allocated by thread T0 here:
    #0 0x7f48252cf0b7 in operator new[](unsigned long) (/opt/compiler-explorer/gcc-11.2.0/lib64/libasan.so.6+0xb30b7)
    #1 0x4015db in main /app/example.cpp:19
    #2 0x7f48243620b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)

SUMMARY: AddressSanitizer: heap-buffer-overflow /app/example.cpp:22 in main

Meaning the arr[i][j]=i; causes causes buffer overflow because of a m<->n typo in the loop for(int j=0; j<m; j++).

user_76917290
Related