如何在C中找到unsigned char指针的长度？

Question

I want to find size or length of an unsigned char pointer in a function where that pointer is an argument to that function.Near the pointers declaration size is coming correctly.

But when i am trying to find size in function it is giving 4.

How can i do this ?

#include <stdio.h>
//void writeFile(unsigned char *da);
void writeFile(char *da);
int main(int arc,char **argv)
{

 unsigned char block_bmp[]=
{
   0x0,0xff,0xff,0xff,0x0,0x0,0x0,0xff,0xff,0xff,0x0,0x0,0x0,0xff,0xff,0xff,//*16bytes*
};
printf("size::%d\n",sizeof(block_bmp));
writeFile(block_bmp);
}

//void writeFile(unsigned char *da)
void writeFile(char *da)
{
printf("%d\n",__LINE__);
printf("size::%d\n",sizeof(da));
printf("length::%d\n",strlen(da));
FILE *fp;
int i;
fp=fopen("/tmp/hexfile","wb");
for(i=0;i<3780;i++)
    fprintf(fp,"%c",da[i]); 
//  fwrite(da,sizeof(unsigned char),sizeof(da),fp);
fclose(fp);
}

Answer 1

If it points to a NULL-terminated string, use strlen . If not, the pointer is just some memory address for the called function, without any additional information about the size of the array (I assume, you try to pass an array). I suggest passing the number of array elements as additional parameter to the function.

Answer 2

You can not use sizeof in this condition. You should be using the strlen if the char array is NULL terminated.

When you pass array to a function, it decays to pointer , you can't use sizeof on this pointer to get the size of the array. The sizeof will give you 4 bytes (assuming 32 bit machine).

Example:

#include <stdio.h>
void fun(int myArray[10])
{
    int i = sizeof(myArray);
    printf("Size of myArray = %d\n", i);
}
int main(void)
{
    // Initialize all elements of myArray to 0
    int myArray[10] = {0}; 
    fun(myArray);
    getch();
    return 0;
}
/**************OUTPUT**************
Size of myArray = 4
***********************************/

Answer 3

unsigned char array[100];
// sizeof(array) == 100

unsigned char* ptr = array;
// sizeof(ptr) == 4 for 32 bit platform.

When you call a function such as

 foo(unsigned char* ptr)
 {
 }

with 'array' as argument, you only see a 'unsigned char *', not an array with 100 elements. That's why 'sizeof' returns 4.

Answer 4

pointers size sizeof(any_pointer_variable) will not depend on the datatype to which it is going to point. The size of the pointer is mostly 4 that is what you are getting in function. and it doesnt depend on what it points to. all pointers will have same size independent of what type they point to.