一本介绍C指针的书--指针和字符串3.2

Of course, what the above program illustrates is a simple way of copying a string. After

playing with the above until you have a good understanding of what is happening, we can

proceed to creating our own replacement for the standard strcpy() that comes with C. It

might look like:

char *my_strcpy(char *destination, char *source)

{

char *p = destination;

while (*source != '\0')

{

*p++ = *source++;

}

*p = '\0';

return destination;

}

In this case, I have followed the practice used in the standard routine of returning a

pointer to the destination.

Again, the function is designed to accept the values of two character pointers, i.e.

addresses, and thus in the previous program we could write:

int main(void)

{

my_strcpy(strB, strA);

puts(strB);

}

I have deviated slightly from the form used in standard C which would have the

prototype:

char *my_strcpy(char *destination, const char *source);

Here the "const" modifier is used to assure the user that the function will not modify the

contents pointed to by the source pointer. You can prove this by modifying the function

above, and its prototype, to include the "const" modifier as shown. Then, within the

function you can add a statement which attempts to change the contents of that which is

pointed to by source, such as:

*source = 'X';

which would normally change the first character of the string to an X. The const modifier

should cause your compiler to catch this as an error. Try it and see.

Now, let's consider some of the things the above examples have shown us. First off,

consider the fact that *ptr++ is to be interpreted as returning the value pointed to by ptr

and then incrementing the pointer value. This has to do with the precedence of the

operators. Were we to write (*ptr)++ we would increment, not the pointer, but that which

the pointer points to! i.e. if used on the first character of the above example string the 'T'

would be incremented to a 'U'. You can write some simple example code to illustrate this.

Recall again that a string is nothing more than an array of characters, with the last

character being a '\0'. What we have done above is deal with copying an array. It happens

to be an array of characters but the technique could be applied to an array of integers,

doubles, etc. In those cases, however, we would not be dealing with strings and hence the

end of the array would not be marked with a special value like the nul character. We

could implement a version that relied on a special value to identify the end. For example,

we could copy an array of positive integers by marking the end with a negative integer.

On the other hand, it is more usual that when we write a function to copy an array of

items other than strings we pass the function the number of items to be copied as well as

the address of the array, e.g. something like the following prototype might indicate:

void int_copy(int *ptrA, int *ptrB, int nbr);

where nbr is the number of integers to be copied. You might want to play with this idea

and create an array of integers and see if you can write the function int_copy() and make

it work.

This permits using functions to manipulate large arrays. For example, if we have an array

of 5000 integers that we want to manipulate with a function, we need only pass to that

function the address of the array (and any auxiliary information such as nbr above,

depending on what we are doing). The array itself does not get passed, i.e. the whole

array is not copied and put on the stack before calling the function, only its address is

sent.

This is different from passing, say an integer, to a function. When we pass an integer we

make a copy of the integer, i.e. get its value and put it on the stack. Within the function

any manipulation of the value passed can in no way effect the original integer. But, with

arrays and pointers we can pass the address of the variable and hence manipulate the

values of the original variables.