Is it possible that in different program executions the code above is not going to work? Yes, it’s about as possible as 1+1==2. Is this correct? Absolutely. How can I solve the problem? Write and read the array rather than its address: vl_size size = vl_get_type_size(dataType); const void *means = vl_gmm_get_means(gmm) out.write(static_cast<const char*>(means), numComponents * … Read more
That is hard to tell, because it depends on the way these pointers were allocated. But, as you have pointers 3 levels deep, I suppose it is not enough to free them. You should have a look at tools like valgrind which helps you to identify memory leaks. To explain a bit, char *** means … Read more
Here’s a brief table of how pointers map to normal (i.e. non-pointer variables) and vice versa (assume you’ve declared the pointer as int *varptr and your regular variable as int v): Dereferencing a pointer: *varptr v Passing the variable by reference: varptr &v Leave a comment if you have further questions and I’ll do my … Read more
Both source and chars are pointers to string literals, you can’t modify a string literal, as per the language rules, the behavior of a program where this is done is undefined. source must be declared as a char array: char source[] = “hello, people.”; This way a copy of the string is stored in the … Read more
There is a big misunderstanding here: You are comparing vectors with different semantics. It’s like asking which is faster, a glass of water or a banana. If you think about using a vector of pointers, std::vector<T*>, you need to be aware that you are using manual memory management. It might be faster in some use-cases, … Read more
OK, I will try to explain this as simply as I can. When you do: putchar(*pch++) what you say is ‘print that character of the address that pch points to and then increment the pch to point to the next address’. Essentially, before the first putchar(), *pch=”a” and after *pch=”b”(because pch points now to ch[1]). … Read more
Change the following statements int ** dp = new int *; *dp = array; Or even you could write one statement instead of the two above int ** dp = new int *( array ); Also do not forget to free the allocated memory delete dp; delete []array; As for your code then you are … Read more
somestruct *mystruct defines a pointer to memory of type somestruct and does not point to anything, or better if it is local variable then it is not initialized and this is Undefined Behaviour. If you do this somestruct mystruct then you define struct itself and not a pointer (object exists in memory). To use pointer … Read more
“How can a pointer to a character data type can hold a string data type?” Well, it’s partly true that in C, type ‘pointer to char’ is the string type. Any function that operates on strings (including printf) will be found to accept these strings via parameters of type char *. “How can printf() print … Read more
head=malloc(sizeof(node*)); Will allocate a space in memory to hold a POINTER to a object of type node. head=malloc(sizeof(node)); Will allocate space for the structure itself, and return to you the address of that memory space, and you will hold that in the head variable For more info, check this other question about Malloc. How do … Read more
You are allocing array of pointers wrong. First you have to allocate enough space for row pointers int** pptr = new int* [rows]; For every pointer enough space for col integers for (int i = 0; i < cols; i++) { pptr[i] = new int [cols]; } To delete arrays use delete[] instead of delete. … Read more
How your code should look #include <stdio.h> #include <stdlib.h> #include <math.h> void myfunction(double x, int y[2], int *d); int main(int argc, char **argv) { int a[2]; int *c = malloc(sizeof(int)); double b=10.0; *c = 5; a[0]=1; a[1]=2; printf(“before call %f %d %d %d\n”,b,a[0],a[1],*c); myfunction(b,a,c); printf(“after call %f %d %d %d\n”,b,a[0],a[1],*c); } void myfunction(double x, int … Read more
As others have said, the reason for the empty space is because you asked it to print out str[3], which contains a space character. Your second question seems to be asking why there’s a difference between printing a char* (it prints the string) and int* (it just prints the address). char* is treated as a … Read more
So you are getting access voilation…. The problem is that your buffers are probably overflowing and you have no length check in your code… The buffers you have allocated are only 25 bytes long char str1[25], str2[25]; so to intruduce a length check, add an extra parameter to concat which tells how long the output … Read more
As we know size of int data type is 2 bytes or 4 bytes depend on your system. While char pointer point one byte at a time. Memory representation of int a = 300; So, char pointer p pointing to only first byte as show above figure. So, first time print 44 output. Then, printf(“%d”,*++p); … Read more