为什么这个C代码（for循环）永远循环？

Question

Learning C right now and I've made this simple for-loop . The program is just to check if the argument given has the 'a' alphabet in it.

int main(int argc, char *argv[])
{
    if(argc != 2) {
        printf("ERROR: Need one argument\n");
        return 1;
    }

    int i = 0;
    char letter;
    for(i = 0, letter = argv[1][i]; letter != '\0'; i++) {

        switch(letter) {
            case 'a':
                printf("%d: 'a'\n", i);
                break;

            default:
                printf("%d: '%c' is not an 'a'\n", i, letter);
        }
    }

    return 0;
}               

The result of this is a forever looping program, but if I change the line:

for(i = 0, letter = argv[1][i]; letter != '\\0'; i++) to

for(i = 0, letter = argv[1][i]; argv[1][i] != '\\0'; i++) ,

The code runs just fine. Why is this?

Answer 1

letter is never updated in your loop. It always have first character of string in argv[1] . Place this statement

letter = argv[1][i];  

after switch statement in the loop body.

Answer 2

letter is assigned value only once; at the beginning of the loop. That assignment should be inside the for loop:

// Assign initial value to letter
char letter = argv[1][0];
for(i = 0; letter != '\0'; i++){
    switch(letter) {
        // ...
    }

    // Update new value to letter
    letter = argv[1][i];
}

Answer 3

The first part of a for loop is only initialized in the beginning. You can "fix" your code by swapping the , and ; there:

for (i = 0; letter = argv[1][i], letter != '\0'; i++) {

Now the condition uses the comma operator to first assign the letter (and discard this value), and then to check that the character is not '\\0'. Actually the 2nd part is superfluous since the value of letter = argv[1][i] is the character, and it is true only if it is not '\\0', so we can write the loop as

for (i = 0; letter = argv[1][i]; i++) {

---

Should you not need a loop index, seasoned C programmer, however, wouldn't use any of the constructs above - including indexing, instead using just a pointer-to-char:

char *pos;
for (pos = argv[1]; *pos; pos++) {
    switch(*pos) {
        case 'a':
            printf("'a'\n");
            break;

        default:
            printf("'%c' is not an 'a'\n", *pos);
    }
}

or even with an index

char letter, *pos = argv[1];
int i;
for (i = 0; letter = pos[i]; i++) {
    switch(letter) {
        case 'a':
            printf("%d: 'a'\n", i);
            break;

        default:
            printf("%d: '%c' is not an 'a'\n", i, letter);
    }
}

Answer 4

If you ever have trouble debugging a for loop, it often helps to think of for (INITIALISE; TEST; INCREMENT) to be equivalent to:

INITIALISE;
while (TEST)
{
    // code
    INCREMENT;
}

Also you should never use the comma operator. It's OK to use it if you really know what you are doing, but if you really knew what you were doing then you wouldn't have posted this question here.

So in your for loop for(i = 0, letter = argv[1][i]; letter != '\\0'; i++) and replacing the comma with a semicolon we can look at it as:

i = 0;
letter = argv[1][i];
while (letter != '\0')
{
    // other code (which doesn't modify letter)
    i++;
}

and from this it should be clear why the loop doesn't terminate.

Answer 5

Whenever you get an accidental for-ever loop, the first thing you check is if you actually increment the loop iterator. You don't, so there you go.

The fix is trivial, just use a pointer instead, which was probably what you intended anyhow?

const char* cptr = argv[1]; 

for(i = 0; cptr[i] != '\0'; i++)
{
  switch(cptr[i])
  { 
    ... 
  }
}

You now have a pure, simple loop. Avoid multiple iterators when possible. As a bonus, it runs faster than the original, since there is no pointless copying at each lap of the loop.