不明白这个for循环是如何工作的

Question

Can someone explain how this loop works? The entire function serves to figure out where in hash to place certain strings and the code is as follows:

//determine string location in hash
int hash(char* str)
{
    int size = 100;
    int sum;

    for(; *str; str++)
        sum += *str;

    return sum % size;
}

It seems to iterate over the string character by character until it hits null, however why does simple *str works as a condition? Why does str++ moves to the next character, shouldn't it be something like this instead: *(str+i) where i increments with each loop and moves "i" places in memory based on *str address?

Answer 1

In C, chars and integers implicitly convert to booleans as: 0 - false, non-zero - true;

So for(; *str; str++) iterates until *str is zero. (or nul)

str is a pointer to an array of chars. str++ increments this pointer to point to the next element in the array and therefore the next character in the string.

So instead of indexing by index. You are moving the pointer.

Answer 2

The condition in a for loop is an expression that is tested for a zero value. The NUL character at the end of str is zero.

The more explicit form of this condition is of course *str != '\\0' , but that's equivalent since != produces zero when *str is equal to '\\0' .

As for why str++ moves to the next character: that's how ++ is defined on pointers. When you increment a char* , you point it to the next char -sized cell in memory. Your *(str + i) solution would also work, it just takes more typing (even though it can be abbreviated str[i] ).

Answer 3

This for loop makes use of pointer arithmetic . With that you can increment/decrement the pointer or add/substract an offset to it to navigate to certain entries in the array, since array are continuous blocks of memory you can do that.

str points to a string. Strings in C always end with a terminating \\0 .

*str dereferences the actual pointer to get the char value.

The for loop's break condition is equivalent to:

*str != '\0'

and

str++

moves the pointer forward to next element.

The hole for-loop is equivalent to:

int len = strlen(str);
int i;
for(i = 0; i < len; i++)
    sum += str[i];

You could also write is as while-loop:

while(*str)
    sum += *str++;

Answer 4

It has to do with how C converts values to "True" and "False". In C, 0 is "False" and anything else is "True"

Since null (the character) happens to also be zero it evaluates to "False". If the character set were defined differently and the null character had a value of "11" then the above loop wouldn't work!

As for the 2nd half of the question, a pointer points to a "location" in memory. Incrementing that pointer makes it point to the next "location" in memory. The type of the pointer is relevant here too because the "Next" location depends on how big the thing being pointed to is

Answer 5

Why does str++ moves to the next character, shouldn't it be something like this
instead: *(str+i) where i increments with each loop and moves "i" places in 
memory based on *str address?

In C/C++, string is a pointer variable that contains the address of your string literal.Initially Str points to the first character.*(str) returns the first character of string.

Str++ points to second charactes.Thus *(str) returns the second character of the string.

why does simple *str works as a condition? 

Every c/c++ string contains null character.These Null Characters signify the end of a character string in C. ASCII code of NUL character is 0 .

In C/C++,0 means FALSE.Thus, NUL Character in Conditional statement 
means FALSE Condition. 

for(;0;)/*0 in conditions means false, hence the loop terminates 
when pointer points to Null Character.
{
}

Answer 6

When the pointer points to a null character it is regarded as false. This happens in pointers. I don't know who defined it, but it happens.

It may be just becuase C treats 0 as false and every other things as true.

For example in the following code.

if(0) {
           puts("true");
} else {
           puts("false"); 
}

false will be the output

Answer 7

The unary * operator is a dereference operator -- *str means "the value pointed to by str ." str is a pointer, so incrementing it with str++ (or ++str ) changes the pointer to point to the next character. So it is the correct way to increment in the for loop.

Any integral value can be treated as a Boolean. *str as the condition of the for loop takes the value pointed to by str and determine if it is non-zero. If so, the loop continues Once it hits a null character, it terminates.