How to print out dash or dot using fprintf/printf?

Question

As of now I'm using below line to print with out dot's

fprintf( stdout, "%-40s[%d]", tag, data);

I'm expecting the output would be something like following,

Number of cards..................................[500]
Fixed prize amount [in whole dollars]............[10]
Is this a high winner prize?.....................[yes]

How to print out dash or dot using fprintf/printf?

Solution 1

A faster approach:

If the maximum amount of padding that you'll ever need is known in advance (which is normally the case when you're formatting a fixed-width table like the one you have), you can use a static "padder" string and just grab a chunk out of it. This will be faster than calling printf or cout in a loop.

static const char padder[] = "......................"; // Many chars

size_t title_len = strlen(title);
size_t pad_amount = sizeof(padder) - 1 - title_len;

printf(title); // Output title

if (pad_amount > 0) {
    printf(padder + title_len); // Chop!
}

printf("[%d]", data);

You could even do it in one statement, with some leap of faith:

printf("%s%s[%d]", title, padder + strlen(title), data);

Solution 2

You can easily do this with iostreams instead of printf

cout << setw(40) << setfill('.') << left << tag[i] << '[' << data[i] << ']' << endl;

Or if you really need to use fprintf (say, you are passed a FILE* to write to)

strstream str;
str << setw(40) << setfill('.') << left << tag[i] << '[' << data[i] << ']' << endl;
printf(%s", str.str());

Solution 3

You can't do it in one statement. You can use sprintf, then substitute dots for spaces yourself, or do something like

int chars_so_far;
char padder[40+1]= '..........'; //assume this is 40 dots.
printf("%.40s%n",tag,&chars_so_far);
printf("%s[%d]",padder+chars_so_far,data);

---

Edit: Simplified my example above based on @Ates' padder concept. This way doesn't require any 'leaps of faith', about whether the tag string is too big or too small - it always starts the data in column 41.

Solution 4

Another solution using a tiny helper function

static inline size_t min(size_t a, size_t b)
{
    return a < b ? a : b;
}

Then, you can do the following:

char padder[] = "........................................";
int len = min(strlen(tag), sizeof(padder) - 1);
printf("%.*s%s[%d]", len, tag, padder + len, data);

This is essentially what Ates posted, but I actually figured this out on my own ;)

Solution 5

You are going to have to output the string with the dot or dash padding yourself.

Something like (forgive my C, it's rusty):

printAmount(char *txt, int amt) {
    printf("%s",txt);
    for(int xa=strlen(txt); xa<40; xa++) { putc('.'); }
    printf("[%d]",amt);
    printf("\n");
    }

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Comments

[Christoph]

your last suggestion breaks if strlen(title) > strlen(padder)

[Ates Goral]

@Christoph, hence the proclaimed "leap of faith" :)

[Christoph]

@Ates: interesting concept, this 'faithful programming' of yours ;)

[Jonathan Leffler]

Change the size_t to int; you must pass an int to the '*' in the printf() family of functions.

[Ates Goral]

Or make the helper more helpful: printf("%s%s[%d]", title, getPadding(title), data);

[Yavor Angelov]

+1 for being easy to read and understand and not breaking under certain (albeit somewhat uncommon situations).

[Ates Goral]

First filling the entire buffer with the pad character, and then overwriting them seems a bit wasteful though...

[Lawrence Dol]

Also, you have a buffer overrun (don't you love C) - s/be: buf[len-1]=0;

[Thi]

is it possible to call a function inside printf?

[Lawrence Dol]

@Thi: Yes this is legitimate - the function is invoked and it's return passed as a parameter to printf... this is normal coding in any procedural, OOP or functional language.

[vitrums]

size_t -> int is required in if ((int)pad_amount > 0)

[sleblanc]

This makes the most sense. If you are padding to a fixed amount of characters, you probably want to chop titles that are too long, so it makes sense to use fixed-width arrays.