How do I convert from void * back to int

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if I have

int a= 5;
long b= 10;
int count0 = 2;
void ** args0;
args0 = (void **)malloc(count0 * sizeof(void *));
args0[0] = (void *)&a;
args0[1] = (void *)&b;

how can I convert from args[0] and args0[1] back to int and long? for example

int c=(something im missing)args0[0]
long d=(something im missing)args1[0]

Assuming that your &a0 and &b0 are supposed to be &a and &b, and that you mean args0[1] for setting up long d, you have stored a pointer to a in args0[0] and a pointer to b in args0[1]. This means you need to convert them to the correct pointer types.

int c = *((int *)args0[0]);
int d = *((long *)args0[1]);

Question about converting `void *` to `int` in C, You're return ing the value of int sum by setting a void * address to it. In this case, the address is not valid. But, if you keep that in mind and get  *i points to a simple integer, while *p is a pointer to a function with no arguments returning an integer. Thus, these two have completely different types, and because you said p shall point to a function returning an int, you can not tell it to point to an int.

To literally answer your question, you'd write

int c = *((int *)args0[0]);
long d = *((long *)args[1]);

What might concern me about your code is that you have allocated space for the pointers to your locations, but you haven't allocated memory for the values themselves. If you expect to persist these locations beyond the local scope, you have to do something like:

int *al = malloc(sizeof(int));
long *bl = malloc(sizeof(long));
*al = a;
*bl = b;
void **args0 = malloc(2 * sizeof(void *));
args0[0] = al;
args0[1] = bl;

Casting, double divide(int a, int b) { double f = a; f /= b; return f; } The key here is to use a pointer of the right type, converting the void pointer value to it  I'm dealing with some code that uses an external library in which you can pass values to callbacks via a void* value.. Unfortunately, the previous person working on this code decided to just pass integers to these callbacks by casting an integer to a void pointer ((void*)val).

Try this:

 int c =  *( (int *)  args0[0]);

 long d = *( (long *) args0[1]);

void* and casts, in C and C++, The reason for this is simple: malloc returns void* and not int*. While in C it's legal to assign void* to int* without a cast, in C++ it isn't. Why the  The type specified before the * in a pointer type is called the referent type. Only an unmanaged type can be a referent type. Pointer types do not inherit from object and no conversions exist between pointer types and object. Also, boxing and unboxing do not support pointers. However, you can convert between different pointer types and between

You need to tell it that the void* should be interpreted as an int* or long* when you dereference.

int a = 5;
long b = 10;
void *args[2];
args[0] = &a;
args[1] = &b;

int c = *(int*)args[0];
long d = *(long*)args[1];

void pointer in C / C++, In C++, we must explicitly typecast return value of malloc to (int *). 2) void pointers in C are used to implement generic functions in C. For example compare  The reason for this is simple: malloc returns void* and not int*. While in C it's legal to assign void* to int* without a cast, in C++ it isn't. Why the difference? Well, let us start with C. The official "bible" of C, "The C Programming Language, 2nd edition" by Kernighan and Ritchie states in section A.6.8: Any pointer to an object may be

While others have answered your question, I will make a comment about the last three lines in the first part of your code snippet:

args0 = (void **)malloc(count0 * sizeof(void *));
args0[0] = (void *)&a;
args0[1] = (void *)&b;

The above is better written as:

args0 = malloc(count0 * sizeof *args0);
args0[0] = &a;
args0[1] = &b;

The malloc() call is easier to read this way, and less error-prone. You don't need a cast in the last two statements since C guarantees conversions to and from an object pointer and a void pointer.

Type conversions - C++ Tutorials, Converting to int from some smaller integer type, or to double from float is A& x) { return * this ;} operator A() { return A();} }; void fn (B x) {} int main () { A foo;  Java String/int FAQ: How do I convert a Java String to an int? Solution. The most direct solution to convert a string to an integer is to use the parseInt method of the Java Integer class. parseInt converts the String to an int, and throws a NumberFormatException if the string can’t be converted to an int type. Here are two short examples.

EXP36-C. Do not cast pointers into more strictly , In this noncompliant code example, loop_function() is passed the char pointer char_ptr but returns an object of type int pointer: int *loop_function(void  In this tutorial you learned how can you convert string into int or number. You also learned to convert string into different data type in c# like decimal, float, double, short, int, long, ushort, uint and ulong with complete programming example. You also learn which type of format can be generated during this type of conversion how can handle

static_cast conversion, 7) Scoped enumeration(C++11) type can be converted to an integer or Conversion of any pointer to pointer to void and back to pointer to the original (or more  A special case of this situation is the function returning void. The keyword "void" does not mean any real type; it simply means the method which returns nothing. Such method is called for its side effect and cannot appear in the assignment operation at all. Therefore, you don't need to "convert" anything — there is nothing to convert.

reinterpret_cast conversion, A pointer converted to an integer of sufficient size and back to the can be converted to any integral type as if it were (void*)0, but no value, not  value2 is void* and in reality points at int but i have to treat it such as i do not know where it points to, but to choose between cases. I can only see the prototype of function. Code:

  • Just as a side remark, I think it is no good practice to cast void* to int or long. The width of integers and of pointers need not necessarily be the same on a given platform. A more portable way is to use intptr_t or even better uintptr_t. This guarantees that you don't loose bits.
  • int c=*((int )args0[0]); causes error: invalid type argument of 'unary *' (have 'int') and int c=((int *)args0[0]); fprintf(stderr,"main(): %d \n",c); causes the output to be -4261808
  • I don't get it. What is the difference between my answer and the one on the top (which is accepted)?
  • When you first posted, the "*" was missing after 'int' for some reason, perhaps due to it being interpreted as formatting -- see w31's comment for how it looked.