We have seen the notation “p=&a; ” used to set the pointer “p” to the address of another variable. For arrays and strings, the technique for setting the pointer to the address of the start of an array or string is a little different.

Consider the following declarations:char s[100];int x[10];char *cptr;int *p;

To set cptr to point to the first element of s[], we can simply say:

cptr = s;This is equivalent to:

cptr = &s[0];

Array and String Addresses

If an array name is used without a subscript, the meaning is “the address of the first element of the array”.

We can set p to point to the start of the x[] array with:p = x;

Once again the statement “p=x;” is the same as “p=&x[0];” The statement “p=&x;” is meaningless and would be syntactically wrong.

A string constant returns the address of the first character of the string. So the statement:

cptr = “This is a string”;

will set cptr to the “address of” the first character of the string, i.e. cptr would point to the character ‘T’ in memory.

Array and String Addresses

Note that the variable cptr does not contain the string (it can’t – cptr is only 4 bytes long and is a pointer), rather the string is placed somewhere in memory by the compiler and the address of the allocated location is assigned to cptr.

Array and String Addresses

The malloc() function is frequently used to request that a block of memory be allocated somewhere in memory and that a pointer be returned to where the allocated space is. Forexample:

char *s;

int *p;s = (char *) malloc(100);p = (int *) malloc(100*sizeof(int));

The argument passed to malloc() is the number of bytes that we want to have allocated. In the first example we are requesting that 100 bytes be allocated and that the pointer variable s be set to point to the space. The function malloc() is, by default, of type void *, meaning that the function returns a pointer but the type of what it points to is undefined.

Dynamically Allocating Space – the malloc() function

By prefixing the call to malloc() with “(char *)” we are type-casting the function’s return value (i.e. overriding the default). This eliminates some compile time warning messages that we would get if we did not do the type-cast.

The second malloc is being used to allocate space for 100 integer values. Once again, remember that the argument to malloc() is the number of bytes to be allocated. Thesizeof(int) function returns the byte size of a single int variable, so this malloc request is requesting 400 bytes of storage.

As with any uninitialized variable or storage area, we can make no assumptions about the initial values in the allocated space.

Dynamically Allocating Space – the malloc() function

We have seen that we can add and subtract values from a pointer. For example, in the function zstrcpy(), we used the pointers src and dst to step through the strings by incrementing each at the end of the loop.

The meaning of doing an arithmetic operation on a pointer is actually a little more involved (and natural) than this example would suggest. Arithmetic operations on pointers are alwaysdone in units related to the size (in bytes) of the object the pointer points to.

Address Arithmetic

For example, consider the following code:

char s[] = “The cat sat on the mat”;int x[] = {1, 2, 3, 4, 5};double f[] = {3.2, 5.43, 6.99, 4.24, 6.11};char *sptr = s;int *xptr = x;double *fptr = f;

Each element of “s” is 1 byte long, each element of “x” is 4 bytes long, and each element of “f” is 8 bytes long (a double precision floating point number occupies 64 bits).

Address Arithmetic

Let’s assume that the first element of s[] is at location 20,000 in memory, the first element of x[] is at location 22,000 in memory and the first element of f[] is at location 24,000. With the above initialization statement, then initially sptr=20,000, xptr=22,000 and fptr=24,000.

Now consider the result of each of the following increment statements:

sptr++;xptr++;fptr++;

In each case, the BYTE length of the data type of what the pointer is pointing to is added. That is, after these increments sptr=20,001, xptr=22,004, and fptr=24,008.

Address Arithmetic

This is fortunate, since this means that the increment positions the pointer to the start of the next element of the array – which is generally just what we want.

This holds for all arithmetic operations on pointers. For example the statement:

xptr+=3;

would actually add “3*4”=”12” to the value of xptr, and will move xptr 3 elements forward in the array.

Address Arithmetic

This program demonstrates(1) how to process command line parameters and(2) how to process an array of ints using a pointer.

/* p20.c */

/* This program illustrates the use of pointers in *//* reading and processing an array of integers *//* It also shows how to access command line args */

/* An upper bound on the number of ints to be read *//* must be specified on the command line.. *//* p2 1000 */

Using a pointer to process an array of numbers

#include <stdio.h>int main(int argc, /* number of command line args */char* argv[]) /* array of pointers to args */{

int* base; // points to start of the arrayint* loc; // array indexint max; // maximum number of values to read inint count; // actual number of values read inint largest; // largest number in the arrayint i;if (argc < 2) /* Make sure at least one arg was given */{

printf("Usage is p20 upper-bound \n");exit(1);

}

Using a pointer to process an array of numbers

The argc parameter contains the number of command line parameters, including the program name itself. Thus a command such as

./a.out 300

will cause argc to be set to 2. It's very important to ensure that the user has provided the number of parameters you need before you attempt to process them!

if (argc < 2) /* Make sure at least one arg was given */{

printf("Usage is p20 upper-bound \n");exit(1);

}

Processing the command line argument

This program expects that a numeric value representing the maximum number of values that are present in the standard input will be present on the command line.

/* The pointer argv[0] points to the name of the program (p20) *//* argv[1] points to the first command line argument *//* The atoi() function converts the ascii character *//* representation an integer to a binary int value. */

max = 0;max = atoi(argv[1]);

if (max <= 0){

printf("upper-bound must be a positive integer \n");exit(2);

}

Processing the command line argument

Note that the size of the area allocated must be specified in bytes. A better way to do this would be to malloc(max * sizeof(int));

count = 0;base = (int *)malloc(4 * max);loc = base;

Allocating storage for the array of ints.

Note that loc and not &loc is passed to scanf(). What would happen if a programmer “accidentally'' passed &loc??

Also note that as each integer is read loc is incremented by 1 and not by 4. The C language automagically takes into account the size of the element pointed to when doing pointerarithmetic! If you were to printf() the value of loc using the p format code, you would see that the actual value does increase by 4 each time it is incremented.

/* Read in the integers from standard input making sure *//* not to overrun the size of the array */while ((scanf("%d", loc) == 1) && (count != max)){

loc = loc + 1;count = count + 1;

}

Reading the input values

Before starting the search for the largest number the value of loc is reset to point to the base of the array.

loc = base; // point loc back to the start of the arraylargest = *loc; // init largest to the first value in the arrayloc = loc + 1;

for (i = 0; i < count; i++){

if (*loc > largest) largest = *loc;

loc = loc + 1;}printf("Largest was %d \n", largest);

}

Identifying the largest value in the array

Exercise: the previous program actually has a nasty bug in it. Use gdb to find and fix it.

Identifying the largest value in the array

The standard runtime library of functions that is normally distributed with C compilers provides a variety of ways to consume command line parameters. In some ways they arebetter than atoi() because they are better at indicating that the user entered incorrect data. Nevertheless, atoi() is probably the most widely used.

Other ways to consume command line parameters

The sscanf() function may be used to attempt to convert ASCII strings in a memory resident buffer to a numeric value. Since argv[1] is a pointer to a memory resident buffer containing the string entered as parameter 1 we could replace the atoi() call in the previous example by:

code = sccanf(argv[1], “%d”, &max);if (code != 1){ fprintf(stderr, “Yeow! bad string in parm 1 \n”);}

Since sscanf() returns the number of values it converted, the variable code will be 1 if it was successful.

The sscanf() function

The strtol() function is more powerful still. It will fill in a pointer to the first illegal character it encounters in the string. If it was successful in producing a valid value, badcharwill point to the NULL character that terminates the string.

char *badchar = NULL;long max = 0;

max = strtol(argv[1], &badchar, 10)if (*badchar != 0){

fprintf(stderr, “Yeow! bad character %c in value\n”, *badchar);

}

The strtol() function

Some data, for example, a grayscale image is most naturally represented as a two dimensional array of the form:

#define NUM_ROWS 768#define NUM_COLS 1024unsigned char pixmap[NUM_ROWS][NUM_COLS];

In this representation each byte represents a single pixel. value of 0 represents completely black and a value of 255 represents the brightest possible white. Intermediate values provide a more or less linear brightness ramp.

To access a specific pixel in such an array one would use:

pixval = pixmap[row][col];

where row and col identify the location of the target pixel within the image.

Representation of multidimensional data

As with all arrays in C, legal values of row range from 0 to NUM_ROWS - 1

A disadvantage of this representation is that the value of NUM_ROWS and NUM_COLS must generally be established at compile time.

We would like to be able to read in the dimensions of the image from the .ppm header and then declare the pixmap array using the actual dimensions of the image.

There is no good way to do that in C because dimensions of static arrays are not allowed to be variables.

Representation of multidimensional data

Thus there are two ways to approach the problem:1. The naive way:

unsigned char pixmap[MAX_ROWS][MAX_COLS];

where MAX_ROWS and MAX_COLS represent the dimensions of the largest image our program is able to handle.

This approach• wastes space• forces us to read or write the image one row at a

time.• constrains the program to work only with images

having size within the predefined limits2. The correct way:

Use a single dimensional malloc'd array and handle the indexing ourselves

Possible approaches to the unknown dimension problem

Suppose the integer variables numrows and numcols represent the number of rows and columns in the image and that they have been correctly set using information in the .ppm header.

Using a single dimension array to represent 2-D data

Space for a grayscale image encoded in binary can be allocated by:

unsigned char* imageloc;imageloc = (unsigned char *)malloc(numrows * numcols);

To read in the grayscale image from the standard input:

pixcount = fread(imageloc, 1, numrows * numcols, stdin);if (pixcount != numrows * numcols){

fprintf(stderr, “pix count err - wanted %d got %d \n”,numrows * numcols, pixcount);exit(1);

}

Grayscale images

In this example we print pixel values of an entire image with one line of output per row of pixel data:

for (row = 0; row < numrows; row = row + 1){

unsigned char* loc;loc = imageloc + row * numcols; // first pix in rowprintf(“\n”);for (col = 0; col < numcols; col = col + 1){

printf(“%03x”, *loc);loc = loc + 1;

}}

Processing an image one row at a time

A color image is often called an rgb image because the red, green, and blue intensities of each pixel are stored together. Space for a color image in binary rgb format can be allocated by:

unsigned char* imageloc;imageloc = (unsigned char *)malloc(3 * numrows * numcols);

To read in the rgb image:

pixcount = fread(imageloc, 3, numrows * numcols, stdin);if (pixcount != numrows * numcols){

fprintf(stderr, “pix count err - wanted %d got %d \n”, numrows * numcols, pixcount);

exit(1);}

Color images

The value of any grayscale pixel at location (row, col) within the image is accessed in the following way:

*(imageloc + row * numcols + col)

or equivalently

imageloc[row * numcols + col]

For example, if the value of numcols is 10, then there are 10 pixels per image row. To reach the pixel whose (row, column) address is (3, 5) it is necessary to pass over three completerows (row 0, row 1, and row 2) and 5 pixels in row three (pixels 0, 1, 2, 3, and 4).

Thus, the offset of the pixel at (3, 5) is 3 * 10 + 5 as shown above.

Accessing a specific element in malloc'd 2-dimensional data

Even though the pixels of a binary grayscale image require one byte of storage and those of a floating point image require four bytes of storage. The value of a floating point pixel at location (row, col) is also

*(imageloc + row * numcols + col)

Why? Because pointer arithmetic automagically takes into account the size of the object pointed to and the C compiler knows that unsigned chars require one byte and floats require 4.

Floating point images

Here the process is slightly grubbier because each pixel is represented by three constituent components (r, g, and b), where each of (r, g, and b) are represented by a single unsigned char. Nevertheless, a bit of reflection yields:

red = *(imageloc + 3 * row * numcols + 3 * col);green = *(imageloc + 3 * row * numcols + 3 * col + 1);blue = *(imageloc + 3 * row * numcols + 3 * col + 2);

It is necessary to multiply by 3 because each pixel occupies three bytes of memory. Since our pointer is a pointer to type unsigned char which is a single byte, there is no magic available from the compiler to help us out here.

Accessing the individual pixels of the binary rgb image