char数组中的整数

Question

I want know how integers are treated in char arrays. For example when we use scanf() & printf() functions for single characters we use "%c" , for string values "%s" , etc. I used "%s" for printing the char array with integers. But it prints some junk characters as the output.

Answer 1

While working with integer variables directly, it's simple as:

int i = 7;
printf("%d", i);

and while working with an array of characters, you could create helper functions that will do the conversion between char* <-> int so that you end up with more portable solution:

int toInt(unsigned char* p) {
    return *(p + 0) << 24 | *(p + 1) << 16 | *(p + 2) << 8 | *(p + 3);
}
...
unsigned char arr[4] = {0,0,0,4};
int i = toInt(&arr[0]);
printf("%d", i);

should print 4

Alternatively you might use simple cast:

unsigned char arr[4] = {0,0,0,4};
int* i = &arr[0];
printf("%d", *i);

but that solution will be depend on endianness. On ideone.com it prints 67108864 .

Answer 2

If you know the length of each phone number will be the same, you could pad out the front of the number with zeros. eg to pad each number out to 10 digits, putting 0's in front when necessary:

int d = 123456789;
printf("%d\n", d); // prints 123456789
printf("%010d\n", d); // prints 0123456789

d = 12345678;
printf("%010d\n", d); // prints 0012345678 :-(

It's hard to guess more about the requirements without seeing the code. There's lots of examples of formatting with printf, eg: http://www.codingunit.com/printf-format-specifiers-format-conversions-and-formatted-output