How many characters is fscanf scanning in my code? - c

Question

I'm using a for loop, which I want to iterate a number of times equal to the number of characters scanned by scanf. However, it seems to run too many times. The code was originally written to print to phrase "We are in 2019", which it does, but now I need to use it for something else. I added in the line: printf("%i",i);
to see how many times it was going through the for loop. It seems to be running 8 times for each run of the while loop, regardless of how long the word scanned was.
#include <stdio.h>
#include <stdlib.h>
int main() {
char* word = malloc(sizeof(char) * (46)); // create a char array
FILE* fp;
fp = fopen("file.txt", "w+");
fputs("We are in 2019", fp);
rewind(fp); // sets to start of file
while(fscanf(fp, "%s", word) != EOF) {
for(int i = 0; i < sizeof(word) / sizeof(word[0]); i++) {
printf("%c", word[i]);
printf("%i", i);
}
printf("\n");
}
fclose(fp);
return (0);
}
The output is:
W0e1234567
a0r1e234567
i0n1234567
200112934567
So I can see it's running the for loop 8 times for each run for each run of the while loop.
Am I misunderstanding how fscanf works? i thought it stopped at a whitespace, and only stored the preceding characters... e.g. first it would scan "We" then store it as a 3 character array in "words", then scan "are" in a 4 character array in "words" and so on. What is really happening?

Answer 1

It is stopping at whitespace.
Apparently you have 64 bit pointers. The variable word is a pointer, so sizeof(word) is 8. Since sizeof one character is 1, sizeof(word) / sizeof(word[0]) is 8.
So, the first time fscanf returns, it has read "We" into the buffer. Then it loops for i from 0 to 7.
i = 0
printf("%c", word[i]); //=> 'W'
printf("%i", i); //=>0
i = 1
printf("%c", word[i]); //=> 'e'
printf("%i", i); //=>1
i = 2
printf("%c", word[i]); //=> '\0' so nothing
printf("%i", i); //=>2
i = 3
printf("%i", i); //=>3
i = 4
printf("%i", i); //=>4
i = 5
printf("%i", i); //=>5
i = 6
printf("%i", i); //=>6
i = 7
printf("%i", i); //=>7
so that is how you get:
W0e1234567
You would see the desired output if your loop looked like this:
while(fscanf(fp, "%s", word) != EOF) {
puts(word);
}

Answer 2

sizeof(word) will not give you the size of the array. It gives the size of the pointer char *word.
It would be simpler to avoid memory allocation, and the size of the array isn't really relevant anyway.
#include <stdio.h>
#include <stdlib.h>
int main() {
char word[46]; // local array
FILE* fp = fopen("file.txt", "w+");
if(fp == NULL) {
return 1; // or better error report
}
fputs("We are in 2019", fp);
rewind(fp);
while(fscanf(fp, "%45s", word) == 1) { // restrict length, check the scan worked
for(int i = 0; word[i] != 0; i++) { // finish when the string terminator is found
printf("%c", word[i]);
}
printf("\n");
}
fclose(fp);
return 0;
}
Program output:
We
are
in
2019

Answer 3

Several remarks can be done on your code
by definition sizeof(char) is 1, so your malloc(sizeof(char) * (46)) can be simplified to be malloc(46). Note you can also use an array char word[46];
fscanf(fp, "%s", word) is very dangerous, if the read word longer than 45 characters you write out of word with an undefined behavior, limit the size doing fscanf(fp, "%45s", word)
As said in the remarks sizeof(word) / sizeof(word[0]) do not return the length of your word, you can just stop when word[i] is 0
to print the index just after each character do not produces a very readable result, are you sure you want that ?
why do you not directly print the word by (f)puts or printf with the format %s and get the length of the string using strlen (if you want to know it) ?
I want to iterate a number of times equal to the number of characters scanned by scanf.
scanf scans more than the number of characters in returns in the string, the spaces are bypassed silently. Only one thing is sure, when you reach the end of the file without error the number of characters scanned by scanf if the size of the file.