R – How to pass an array of struct using pointer in c/c++

cparametersstruct

in C code I'm stuck to pass an array of struct to a function, here's the code that resembles my problem:

typedef struct
{
   int x;
   int y;
   char *str1;
   char *str2;
}Struct1;

void processFromStruct1(Struct1 *content[]);
int main()
{
    Struct1 mydata[]=
    { {1,1,"black","cat"},
      {4,5,"red","bird"},
      {6,7,"brown","fox"},
    };

    processFromStruct1(mydata);//how?!?? can't find correct syntax

    return 0;
}

void processFromStruct1(Struct1 *content[])
{
    printf("%s", content[1]->str1);// if I want to print 'red', is this right?
        ...
}

Compile error in msvc is something like this:

error C2664: 'processFromStruct1' : cannot convert parameter 1 from 'Struct1 [3]' to 'Struct1 *[]'
1>       Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast

How to solve this? tnx.

Best Answer

You almost had it, either this

void processFromStruct1(Struct1 *content);

or this

void processFromStruct1(Struct1 content[]);

and, as Alok points out in comments, change this

content[1]->str1

to this

content[1].str1

Your array is an array of structures, not an array of pointers, so once you select a particular structure with [1] there is no need to further dereference it.

Related Solutions

How to pass a function as a parameter in C

Declaration

A prototype for a function which takes a function parameter looks like the following:

void func ( void (*f)(int) );

This states that the parameter f will be a pointer to a function which has a void return type and which takes a single int parameter. The following function (print) is an example of a function which could be passed to func as a parameter because it is the proper type:

void print ( int x ) {
  printf("%d\n", x);
}

Function Call

When calling a function with a function parameter, the value passed must be a pointer to a function. Use the function's name (without parentheses) for this:

func(print);

would call func, passing the print function to it.

Function Body

As with any parameter, func can now use the parameter's name in the function body to access the value of the parameter. Let's say that func will apply the function it is passed to the numbers 0-4. Consider, first, what the loop would look like to call print directly:

for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
  print(ctr);
}

Since func's parameter declaration says that f is the name for a pointer to the desired function, we recall first that if f is a pointer then *f is the thing that f points to (i.e. the function print in this case). As a result, just replace every occurrence of print in the loop above with *f:

void func ( void (*f)(int) ) {
  for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
    (*f)(ctr);
  }
}

Source

C++ – How to set, clear, and toggle a single bit

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.