Processing when the number of numerical elements in the C language file is not determined for each line

Read a string line by line

Get file data line by line using fgets function

main.c

char data[80];
unsigned int i;
    
for(i = 0; i < N && fgets(data, sizeof(data), fp) != NULL; i++){
    //This will store the string on each line in data
    //At this point, the string contains spaces
    printf("data [%d] %s\n", i, data);
}

/*output
data [0] 10 20 30 40 50
data [1] 12 3  4    5  6
:
data [N] ???
*/

Stores space-separated strings in an array (+ converts to a number)

Get numbers separated by whitespace using strtok function The strtok function passes the first pointer of the string first, then NULL the second and subsequent times. Type conversion from char to int with atoi function

main.c

char data[80];
int num[N][80] = {{0}};  //The numbers in the file are stored here
char *te, *ee;
unsigned int i, j, k;
int temp = 0;
    
for(i = 0; i < N && fgets(data, sizeof(data), fp) != NULL; i++){

    te = data; //String pointer
    j = 0;
    //A character is entered in ee, and the end of the character is'\0'Replace with
    while((ee = strtok(te, " \n")) != NULL) { 
        num[i][j] = atoi(ee); //Convert to number and store in array
        te = NULL; //strtok()Store NULL for the second and subsequent calls to
        j++;
    }
}

Numerical processing

You can manipulate the numerical value by playing with num as follows. Since the number of elements is not acquired, only data at a specific position, maximum value, minimum value, linear search, etc. can be performed. If you can get the number of elements well in the above stage, you may be able to get the average and standard deviation in the data row by row. If you stop at the point of 0, you can get the number of elements itself, but this time it is abandoned because there is 0 in the range of integer values.

main.c

//Display only the first integer value in the i-th row of the array
printf("%d\n", num[i][0]);

//Show maximum value in row i of array
temp = num[i][0];
for(k = 1; k < (sizeof(num[i]) / sizeof(int)); k++){
    if(temp < num[i][k]) {
        temp = num[i][k];
    }
}
printf("max = %d \n", temp);

Whole code

main.c

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 10000

int main() {
    FILE *fp;
    char data[80];
    int num[N][80] = {{0}};
    char *te, *ee;
    unsigned int i, j, k;
    int temp = 0;
    

    if((fp = fopen("./input.txt", "r")) == NULL) exit(1);

    for(i = 0; i < N && fgets(data, sizeof(data), fp) != NULL;i++){
        te = data;
        j = 0;
        while((ee = strtok(te, " \n")) != NULL){
            num[i][j] = atoi(ee);
            j++;
            te = NULL;
        }

        printf("first = %d\n", num[i][0]);

        temp = num[i][0];
        for(k = 0; k < (sizeof(num[i]) / sizeof(int)); k++){
            printf("%d ", num[i][k]);     //Show all
            if(temp < num[i][k]){
                temp = num[i][k];
            }
        }
        printf("\n");
        printf("max = %d \n", temp);
        //printf("\n");
    }
    fclose(fp);

    return 0;
}

Disadvantages, points to keep in mind, etc.

--Since the array has a fixed length, 0 is entered except for the entered numbers. (In this case, the number of arrays is fixed at 80) ← Dynamic array? Should I do it? --As a process, you cannot find anything other than searching for the maximum and minimum values. ← If you can get a good number of elements, you can get an average etc. --If there is a certain rule in the numerical sequence of input.txt, read it with fscanf (). ――Usually, the number of elements is fixed or the order of data is fixed, so there may be almost no such situation. ――In the first place, is it possible to count the number of elements by looking at the line breaks on any line and separating the elements and spaces? (I do not know the method)

Addendum / Acknowledgments

In the comments below

--To @fujitanozomu for ideas on getting the number of elements in each row --@ angel_p_57's implementation idea using strtol ()

I told you! informative. Thank you