C - 2D Array Sgmentation Fault

Question

I have the following code:

int width = 10;
int height = 9;

float** matrix = (float**) malloc(height*sizeof(float));

for (unsigned int i = 0; i < height; i++)
    matrix[i] = (float*) malloc(width*sizeof(float));

//This works
matrix[6][0] = 3.0f;

for (unsigned int j = 0; j < width; j++) {
    for (unsigned int i = 0; i < height; i++) {
        //This fails on [6][0]
        matrix[i][j] = 3.0f;
    }
}

I am attempting to create a 2D array and initialise it however a segmentation fault is received when setting the value of matrix[6][0] within the loop. What I'm finding very strange is that no error is thrown when I set matrix[6][0] outside of the loop. It's my understanding that a segmentation fault occurs when illegal memory is accessed but I cannot find any reason why different memory is being accessed within the loop. I have even examined the assembly code to find out what's happening but I cannot find the problem.

Update: The code was part of a CUDA program (C++) but my brain was half thinking about C and half about C++, hence why it became a mess.

Note that it is [redundant and potentially dangerous to cast the result of malloc and friends in C](http://stackoverflow.com/q/605845/253056). (Doubly so in the above code!) — Paul R, Mar 30 '16 at 16:52
There is no 2D array and nothing which can be used as one in your code. A pointer is not an array! And your wildly casting for the first `malloc` is a clear signal you lack some fundamental basics about pointers, too - no offence (honestly!), just a fact. — too honest for this site, Mar 30 '16 at 16:54
Also C and C++ are different languages. Don't use both tags if the question is about one of them only. If in doubt, always tag for the compiler you use. — too honest for this site, Mar 30 '16 at 16:58
Thanks @Olaf - the code was part of a CUDA program (C++) but my brain was half thinking about C and half about C++, hence why it became a mess. — John Wheal, Mar 30 '16 at 17:17

score 2 · Accepted Answer · answered Mar 30 '16 at 16:48

2

float** matrix = (float**)(float*) malloc(height*sizeof(float));

Instead of sizeof(float) use sizeof(float *) . You need to allocate memory for height number of float pointers .

So your code goes like this -

float** matrix = malloc(height*sizeof(float *));

Your code produces seg fault while using loop as you don't correct amount of memory for float pointers and you end up accessing invalid memory , thus getting undefined behaviour.

Note- Don't use cast for malloc . And check retrun of malloc.

answered Mar 30 '16 at 16:48

ameyCU

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Isn't the size of a float larger than he size of a pointr? If yes, then that cannot cause the segmentation fault. – Paul Ogilvie Mar 30 '16 at 16:50
2

@PaulOgilvie: on a 64 bit OS a `float *` pointer is 8 bytes whereas a `float` is typically only 4 bytes. QED. – Paul R Mar 30 '16 at 16:53
@QED, thanks for the info. So it did cause the seg fault. – Paul Ogilvie Mar 30 '16 at 16:54

score 2 · Answer 2 · answered Mar 30 '16 at 16:50

2

Here is the refined code:

int width = 10;
int height = 9;

float **matrix = malloc(height * sizeof (float *));

for (size_t i = 0; i < height; i++)
    matrix[i] = malloc(width * sizeof(float));

matrix is "a pointer to pointers to floats", while matrix[i] are "pointers to floats".

answered Mar 30 '16 at 16:50

nalzok

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That won't compile it says "error: a value of type "void *" cannot be used to initialize an entity of type "float **" – John Wheal Mar 30 '16 at 16:58
@JohnWheal That must be because you maybe compiling it with C++ compiler . For C these cast are not needed . – ameyCU Mar 30 '16 at 17:02
@ameyCU This is actually part of a CUDA program and so it is being compiled with a C++ compiler. Does this change things? – John Wheal Mar 30 '16 at 17:05
@JohnWheal Then maybe you need to apply cast . Or use `new` and `delete` . – ameyCU Mar 30 '16 at 17:08

C - 2D Array Sgmentation Fault

2 Answers2