fputc(int, FILE *) problem

I'm trying to write a program that inserts a number line into a text file.
For some reason, fputc(int, FILE *) doesn't print out the int to stdout.

Code:

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#include <stdio.h>

int main(int argc, char **argv)
{
        FILE *ifp = fopen(argv[1], "r");
        char c;

        int lineNumber = 0;
        while((c = getc(ifp)) != EOF)
        {
                fputc(c, stdout); //print char to screen
                if(c == '\n')
                {
                        fputc(lineNumber, stdout);
                        lineNumber++;
                }
        }
        printf("\nputc() ran %i times.\n\n", lineNumber);



        fclose(ifp);

        return 0;
}

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input file

Code:

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a
b
c
d

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output

Code:

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a
b
c
d

putc() ran 4 times.

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stdout is line-buffered. meaning characters arent guaranteed to be printed even after a print statement. to ensure things are printed you need to force it to print, for example printing a '\n' newline character after your fputc statements.

I'm not sure what you mean. I updated my code to:

Code:

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        int lineNumber = 0;
        while((c = getc(ifp)) != EOF)
        {
                fputc(c, stdout); //print char to screen
                fputc('\n', stdout);
                if(c == '\n')
                {
                        fputc(lineNumber, stdout);
                        fputc('\n', stdout);
                        lineNumber++;
                }
        }
        printf("\nputc() ran %i times.\n\n", lineNumber);

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New Output:

Code:

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[fl76a06@cisweb ex04]$ ./ex04 test
a



b



c



d




putc() ran 4 times.

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It prints the newlines just fine, but no ints. I suspected this was some sort of buffer problem. Am I putting the calls to putc('\n', stdout) in the correct places?

Code:

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fputc(lineNumber, stdout)

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And you expect ASCII character 1 to be printable, why? (Although if you had went to g, you should have gotten a beep for your troubles.) If you want to print a number, print a number.

when i saw your output it didnt look like anything else was being printed which is why i suggested about the buffering or printing a newline character.

when i tried your first code snippet it was printing something (garbled ASCII). so it doesnt look like its any buffer-mode problem.

notice that your trying to call fputc on an integer, lineNumber. fputc is to print a single character, that is, 1byte. an integer is not 1 byte--its usually 4. so it was printing your 'integer', but only the 1st byte of it, so its not really recognizable.

you can either leave your lineNumber as an int and initialize it to '0' (as opposed to just 0). when you print it in the last printf statement youll have to print it as a %c, not as an integer.

setting the int to '0' will set it to the ASCII value of the zero character. you can leave the ++ statement there because it will simply increase it to the next ASCII character, '1'. however, if you loop more than 10 times, your integer wont represent a number, itll represent the next ASCII value after the character '9'. an easier solution is to simply use printf or fputs/fprint.

You're trying to print 0x00 to 0x03 which aren't printable ASCII characters.
If the value of the int isn't a printable ASCII character you'll get all kinds of crap on the screen. Maybe you want to convert the int to a char first?
or you could just use:

Code:

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printf( "%d", lineNumber );

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Quote:

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Originally Posted by nadroj

notice that your trying to call fputc on an integer, lineNumber. fputc is to print a single character, that is, 1byte. an integer is not 1 byte--its usually 4. so it was printing your 'integer', but only the 1st byte of it, so its not really recognizable.

you can either leave your lineNumber as an int and initialize it to '0' (as opposed to just 0). when you print it in the last printf statement youll have to print it as a %c, not as an integer.

setting the int to '0' will set it to the ASCII value of the zero character. you can leave the ++ statement there because it will simply increase it to the next ASCII character, '1'. however, if you loop more than 10 times, your integer wont represent a number, itll represent the next ASCII value after the character '9'. an easier solution is to simply use printf or fputs/fprint.

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Of course it's recognizable, just not the way the OP intended. putc() takes an int parameter, but only uses the lower 8 bits, so if the int was over 255 it wouldn't see the whole thing.

Initializing the int to '0' won't help unless the file doesn't have more than 10 lines.

As for the printf(), you should NOT use %c since that'll just have the same problem as the putc().

Quote:

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Of course it's recognizable, just not the way the OP intended.

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well no, its not recognizable. because printing the LSB of the integer seems random--ie not recognizable.

Quote:

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Initializing the int to '0' won't help unless the file doesn't have more than 10 lines.

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thats what i mentioned above--it will work if its in the range 0-9 inclusive.

Quote:

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As for the printf(), you should NOT use %c since that'll just have the same problem as the putc().

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using %c will work, when simply printing to the screen as the poster was doing. it will print the value of lineNumber, which is an integer representation of the single character 0-9.

Thanks guys.

Yes, but what I'm saying is, why limit your program to only 10 lines when you could easily use printf( "%d", lineNumber ); and won't have a problem unless your file has over 4294967295 lines?

yes im aware. i agree that using what the OP propsed above is not the best way, but i only focused on the problem given, as times there are requirements to meet, or just someone wants to try something different. above i also suggested to use a different print method that allows to print the actual integer.

Quote:

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Originally Posted by cpjust

Yes, but what I'm saying is, why limit your program to only 10 lines when you could easily use printf( "%d", lineNumber ); and won't have a problem unless your file has over 4294967295 lines?

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Actually, 2147483647 lines -- assuming a signed int. ;)