How to read a multi line using fscanf - c

Question

I want to read my data.txt file which looks like and store it in an array called buffer[i][j]
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
I am writing a code which looks like
#include"stdio.h"
#include"stdlib.h"
int main() {
FILE *fp1;
int i,j;
int buffer[4][4]={0};
fp1 = fopen("exact_enumerated_config_442_cub_mc","r");
for(i=0;i<4;i++) {
for(j=0;j<4;j++) {
fscanf(fp1,"%d", &buffer[i][j]);
}
// fscanf(fp1,"\n");
}
fclose(fp1);
for(i=0;i<4;i++) {
for(j=0;j<4;j++) {
printf("%d ",buffer[i][j]);
}
printf("\n");
}
}
but i get the output...
1 1 2 1
5 1 6 1
17 1 18 1
21 1 22 1
why????

Answer 1

Always check the result of fopen() to ensure the file has been opened.
Always check the result of fscanf() to ensure it was successful and prevent subsequent code processing variables that may not have been assigned a value (it returns the number of assignments made).
Add a leading space character to the format specifier to skip whitespace, including newline characters: " %d".
The code will treat a single line with sixteen ints the same as four lines with four ints. If it is important that the format of the file is four ints per line then read a single line using fgets() and then use sscanf() to extract the ints with the %n format specifier to ensure full buffer was processed:
int ints[4][4] = { { 0 } };
char buffer[1024];
for (int i = 0; i < 4 && fgets(buffer, 1024, fp); i++)
{
int pos;
if (sscanf(buffer,
"%d %d %d %d%n",
&ints[i][0],
&ints[i][1],
&ints[i][2],
&ints[i][3],
&pos) != 4 || pos != strlen(buffer) - 1)
{
fprintf(stderr, "Invalid format: <%s>\n", buffer);
exit(1);
}
}

Answer 2

Add space at the beginning of the string format it should be " %d" to avoid the newline problems
fscanf(fp1," %d", &buffer[i][j]);
BTW you could use the following code instead
for(i=0;i<4;i++) {
fscanf(fp1," %d %d %d %d", &buffer[i][0], &buffer[i][1], &buffer[i][2], &buffer[i][3]);
}

Answer 3

To get required output (1111,2222,...) change:
fp1 = fopen("exact_enumerated_config_442_cub_mc","r");
to:
fp1 = fopen("data.txt","r");
Clarification: when using fopen you should write the name of file you want to read. In you case you have to write data.txt, not exact_enumerated_config_442_cub_mc...
There is no file with this name, moreover there is no any data like 1 1 1 1, 2 2 2 2, 3 3 3 3, 4 4 4 4...
For more detailes visit:
wikibooks.org/wiki/C_Programming/C_Reference/stdio.h/fopen
Here is your "modified" (excess/waste/extra { } removed and data.txt is written) code that gives you required output: 1 1 1 1, 2 2 2 2, 3 3 3 3, 4 4 4 4.
It prints the array named buffer. It means that...
...data was successfully copied! from "data.txt" to buffer[4][4]:
#include"stdio.h"
#include"stdlib.h"
int main()
{
FILE *fp1;
int i,j;
int buffer[4][4];
for ( i = 0; i < 4; i++ )
for ( j = 0; j < 4; j++ )
buffer[i][j] = 0;
fp1 = fopen("data.txt","r");
for(i=0; i<4; i++)
for(j=0; j<4; j++)
fscanf(fp1,"%d", &buffer[i][j]);
fclose(fp1);
for(i=0; i<4; i++)
{
for(j=0; j<4; j++)
printf("%d ",buffer[i][j]);
printf("\n");
}
return 0;
}
P.S.
If data.txt will contain not
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
but
1 1 1 1
2 2 2 2
3 3 3 3
4 444 // the last two elements are absent
the program will read 1-st, 2-nd and 3-rd line properly, and the output of the 4-th line will be
4 444 0 0
It prints 4, then 444, and then 0 and 0: the last two elements are '0's because buffer had been initialized by zeros, so all elements changed their values, but the last two remained to be zeros.

Answer 4

You have said you want to read data.txt then why are you opening the file exact_enumerated_config_442_cub_mc
Try changing this
fp1 = fopen("exact_enumerated_config_442_cub_mc","r");
to
fp1 = fopen("data.txt","r");