How does "while(*s++ = *t++)" copy a string?

My question is, what does this code do (from http://www.joelonsoftware.com/articles/CollegeAdvice.html):

while (*s++ = *t++);

the website says that the code above copies a string but I don't understand why...

does it have to do with pointers?

It is equivalent to this:

while (*t) {
    *s = *t;
    s++;
    t++;
}
*s = *t;

When the char that t points to is '\0', the while loop will terminate. Until then, it will copy the char that t is pointing to to the char that s is pointing to, then increment s and t to point to the next char in their arrays.

This has so much going on under the covers:

while (*s++ = *t++);

The s and t variables are pointers (almost certainly characters), s being the destination. The following steps illustrate what's happening:

• the contents of t (*t) are copied to s (*s), one character.
• s and t are both incremented (++).
• the assignment (copy) returns the character that was copied (to the while).
• the while continues until that character is zero (end of string in C).

Effectively, it's:

while (*t != 0) {
    *s = *t;
    s++;
    t++;
}
*s = *t;
s++;
t++;

but written out in a much more compact way.

Let's assume s and t are char *s that point to strings (and assume s is at least as large as t). In C, strings all end in 0 (ASCII "NUL"), correct? So what does this do:

*s++ = *t++;

First, it does *s = *t, copying the value at *t to *s. Then, it does s++, so s now points to the next character. And then it does t++, so t points to the next character. This has to do with operator precedence and prefix vs. postfix increment/decrement.

Operator precedence is the order in which operators are resolved. For a simple example, look:

4 + 2 * 3

Is this 4 + (2 * 3) or (4 + 2) * 3? Well, we know it is the first one because of precedence - the binary * (multiplication operator) has higher precedence than the binary + (addition operator), and is resolved first.

In *s++, we have unary * (pointer dereference operator) and unary ++ (postfix increment operator). In this case, ++ has higher precedence (also said to "bind tighter") than *. If we had said ++*s, we would increment the value at *s rather than the address pointed to by s because prefix increment has lower precedence* as dereference, but we used postfix increment, which has higher precedence. If we had wanted to use prefix increment, we could have done *(++s), since the parenthesis would have overridden all lower precedences and forced ++s to come first, but this would have the undesirable side effect of leaving an empty character at the beginning of the string.

Note that just because it has higher precedence doesn't mean it happens first. Postfix increment specifically happens after the value has been used, which his why *s = *t happens before s++.

So now you understand *s++ = *t++. But they put it in a loop:

while(*s++ = *t++);

This loop does nothing - the action is all in the condition. But check out that condition - it returns "false" if *s is ever 0, which means *t was 0, which means they were at the end of the string (yay for ASCII "NUL"). So this loop loops as long as there are characters in t, and copies them dutifully into s, incrementing s and t all the way. When this loop exits, s has been NUL-terminated, and is a proper string. The only problem is, s points to the end. Keep another pointer handy that points to the beginning of s (i.e. s before the while() loop) - that will be your copied string:

char *s, *string = s;
while(*s++ = *t++);
printf("%s", string); // prints the string that was in *t

Alternatively, check this out:

size_t i = strlen(t);
while(*s++ = *t++);
s -= i + 1;
printf("%s\n", s); // prints the string that was in *t

We started by getting the length, so when we ended, we did more pointer arithmetic to put s back at the beginning, where it started.

Of course, this code fragment (and all my code fragments) ignore buffer issues for simplicity. The better version is this:

size_t i = strlen(t);
char *c = malloc(i + 1);
while(*s++ = *t++);
s -= i + 1;
printf("%s\n", s); // prints the string that was in *t
free(c);

But you knew that already, or you'll soon ask a question on everyone's favorite website about it. ;)

* Actually, they have the same precedence, but that's resolved by different rules. They effectively have lower precedence in this situation.

while(*s++ = *t++);

Why do people think it is equivalent to:

while (*t) {
    *s = *t;
    s++;
    t++;
}
*s = *t; /* if *t was 0 at the beginning s and t are not incremented  */

when it obviously isn't.

char tmp = 0;
do {
   tmp = *t;
   *s = tmp;
   s++;
   t++;
} while(tmp);

is more like it

EDIT: Corrected a compilation error. The tmp variable must be declared outside of the loop.

The aspect that is mysterious about this is the order of operations. If you look up the C language spec, it states that in this context, the order of operations is as follows:

1. * operator
2. = (assignment) operator
3. ++ operator

So the while loop then becomes, in english:

while (some condition):
Take what is at address "t" and copy it over to location at address "s".
Increment "s" by one address location.
Increment "t" by one address location.

Now, what is "some condition"? The C lang specification also says that the value of an assignment expression is the assigned value itself, which in this case is *t.

So "some condition" is "t points to something that is non-zero", or in a simpler way, "while the data at location t is not NULL".

The C Programming Language (K&R) by Brian W. Kernighan and Dennis M. Ritchie gives a detailed explanation of this.

Second Edition, Page 104:

> 5.5 Character Pointers and Functions > > A string constant, written as > "I am a string" > > is an array of characters. In the internal representation, the array is terminated with the null character '\0' so that programs can find the end. The length in storage is thus one more than the number of characters between the double quotes. > > Perhaps the most common occurrence of string constants is as arguments to functions, as in > printf("hello, world\n"); > > Where a character string like this appears in a program, access to it is through a character pointer; printf receives a pointer to the beginning of the character array. That is, a string constant is accessed by a pointer to its first element. > > String constants need not be functions arguments. If pmessage is declared as > char pmessage; > > then the statement > pmessage = "now is the time"; > > assigns to pmessage a pointer to the character array. This is not a string copy; only pointers are involved. C does not provide any operators for processing an entire string of characters as a unit. > > There is an important different between these definitions: > char amessage[] = "now is the time"; / an array */ char pmessage = "now is the time"; / a pointer / > > amessage is an array, just big enough to hold the sequence of characters and '\0' that initializes it. Individual characters within the array may be changed by amessage will always refer to the same storage. On the other hand, pmessage is a pointer, initialized to point to a string constant; the pointer may subsequently be modified to point elsewhere, but the result is undefined if you try to modify the string contents. > +---+ +--------------------+ pmessage: | o-------->| now is the time \0 | +---+ +--------------------+ > +--------------------+ amessage: | now is the time \0 | +--------------------+ > > We will illustrate more aspects of pointers and arrays by studying versions of two useful functions adapted from the standard library. The first function is strcpy(s,t), which copies the string t to the string s. It would be nice just to say s = t but this copies the pointer, not the characters.To copy the characters, we need a loop. The array version is first: > / strcpy: copy t to s; array subscript version */ void strcpy(char *s, char t) { int i; >
i = 0; while((s[i] = t[i]) != '\0') i ++; } > > For contrast, here is a version of strcpy with pointers: > / strcpy: copy t to s; pointer version 1 */ void strcpy(char *s, char *t) { while((*s = t) != '\0') { s ++; t ++; } } > > Because arguments are passed by value, strcpy can use the parameters s and t in any way it pleases. Here they are conveniently initialized pointers, which are marched along the arrays a character at a time, until the '\0' that terminates t has been copied to s. > > In practice, strcpy would not be written as we showed it above. Experienced C programmers would prefer > / strcpy: copy t to s; pointer version 2 */ void strcpy(char *s, char *t) { while((*s++ = t++) != '\0') ; } > > This moves the increment of s and t into the test part of the loop. The value of *t++ is the character that t pointed to before t was incremented; the postfix ++ doesn't change t until after this character has been fetched. In the same way, the character is stored into the old s position before s is incremented. This character is also the value that is compared against '\0' to control the loop. The net effect is that characters are copied from t to s, up to and including the terminating '\0'. > > As the final abbreviation, observe that a comparison against '\0' is redundant, since the question is merely whether the expression is zero. So the function would likely be written as > / strcpy: cope t to s; pointer version 3 */ void strcpy(char *s, char *t) { while(*s++ = *t++); } > > Although this may seem cryptic as first sight, the notational convenience is considerable, and the idiom should be mastered, because you will see if frequently in C programs. > > The strcpy in the standard library (<string.h>) returns the target string as its function value.

This is the end of the relevant parts of this section.

PS: If you enjoyed reading this, consider buying a copy of K&R - it is not expensive.

It works by copying characters from the string pointed to by 't' into the string pointed to by 's'. For each character copies, both pointers are incremented. The loop terminates when it finds a NUL character (equal to zero, hence the exit).

HINTS:

• What does the operator '=' do?
• What is the value of the expression "a = b"? Eg: if you do "c = a = b" what value does c get?
• What terminates a C string? Does it evaluate true or false?
• In "*s++", which operator has higher precedence?

ADVICE:

• Use strncpy() instead.

Yes, it does have to do with pointers.

The way to read the code is this: "the value that is pointed to by the pointer "s" (which gets incremented after this operation) gets the value which is pointed to by the pointer "t" (which gets incremented after this operation; the entire value of this operation evaluates to the value of the character copied; iterate across this operation until that value equals zero". Since the value of the string null terminator is the character value of zero ('/0'), the loop will iterate until a string is copied from the location pointed to by t to the location pointed to by s.

it copies a string because arrays are always passed by reference, and string is just a char array. Basically what is happening is (if i remember the term correctly) pointer arithmetic. Here's a bit more information from wikipedia on c arrays.

You are storing the value that was dereferenced from t in s and then moving to the next index via the ++.

Say you have something like this:

char *someString = "Hello, World!";

someString points to the first character in the string - in this case 'H'.

Now, if you increment the pointer by one:

someString++

someString will now point to 'e'.

while ( *someString++ );

will loop until whatever someString points at becomes NULL, which is what signals the end of a string ("NULL Terminated").

And the code:

while (*s++ = *t++);

is equal to:

while ( *t != NULL ) { // While whatever t points to isn't NULL
	*s = *t;           // copy whatever t points to into s
	s++;
	t++;
}

Many adherents of С language are convinced that the "while (* s ++ = * t ++)" is a genuine grace.

In the conditional expression of the loop "while",three side effects are inserted(shift of one pointer, shift of the second pointer, assignment).

The body of the loop as a result was empty, since all the functionality is placed in a conditional expression.

1. use for with int i:

	char t[]="I am a programmer",s[20];
	for(int i=0;*(t+i)!='\0';i++)
		*(s+i)=*(t+i);
	*(s+i)=*(t+i); //the last char in t '\0'
	printf("t is:%s\n",t);
	printf("s is:%s\n",s);

2. use for with pointer++:

	char t[]="I am a programmer",s[20];
	char *p1,*p2;
	p1=t,p2=s;
	for(;*p1!='\0';p1++,p2++)
		*p2 = *p1;
	*p2 = *p1;
	printf("t is:%s\n",t);
	printf("s is:%s\n",s);

3. use while with pointer++:

	char t[]="I am a programmer",s[20];
	char *p1,*p2;
	p1=t,p2=s;
	while(*p2++=*p1++);
	printf("t is:%s\n",t);
	printf("s is:%s\n",s);
	printf("t is:%s\n",p1-18);
	printf("s is:%s\n",p2-18);

4. use array to initialize pointers:

	char a[20],*t="I am a programmer",*s;
	s=a;
	while(*s++=*t++);
	printf("t is:%s\n",t-18);
	printf("s is:%s\n",s-18);
	printf("s is:%s\n",a);

Yes this uses pointers, and also does all the work while evaluating the while condition. C allows conditional expressions to have side-effects.

The "*" operator derefereces pointers s and t.

The increment operator ("++") increments pointers s and t after the assignment.

The loop terminates on condition of a null character, which evaluates as false in C.

One additional comment.... this is not safe code, as it does nothing to ensure s has enough memory allocated.

starts a while loop....

*s = *t goes first, this assigns to what t points at to what s points at. ie, it copies a character from t string to s string.

what is being assigned is passed to the while condition... any non zero is "true" so it will continue, and 0 is false, it will stop.... and it just happens the end of a string is also zero.

s++ and t++ they increment the pointers

and it all starts again

so it keeps assigning looping, moving the pointers, until it hits a 0, which is the end of the string

The question I provided the following answer on was closed as a duplicate of this question, so I am copying the relevant part of the answer here.

The actual semantic explanation of the while loop would be something like:

for (;;) {
    char *olds = s;             // original s in olds
    char *oldt = t;             // original t in oldt
    char c = *oldt;             // original *t in c
    s += 1;                     // complete post increment of s
    t += 1;                     // complete post increment of t
    *olds = c;                  // copy character c into *olds
    if (c) continue;            // continue if c is not 0
    break;                      // otherwise loop ends
}

The order that s and t are saved, and the order that s and t are incremented may be interchanged. The save of *oldt to c can occur any time after oldt is saved and before c is used. The assignment of c to *olds can occur any time after c and olds are saved. On the back of my envelop, this works out to at least 40 different interpretations.

Well this is true just in the case of the char if there is no \0 and the it is an integer array the the program will crash because there will be a address whose elements are not the part of the array or pointer, if the system has memory that was allocated using the malloc then the system will keep giving the memory