What is the purpose of this program ? (when EOF means end of file. It's a sign* that the end of a file is reached, and that there will be no data anymore)#include void main() {FILE *ptr;char c, a;ptr = fopen("E:\\exam.txt" , " r+");int N=0;%3Ddo {fscanf(ptr,"%c", &c);if ( c=='\n'){printf("%d\n",N);N=0}N++;}while (c != EOF);fclose (ptr);

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What is the purpose of this program ? (when EOF means end of file. It's a sign * that the end of a file is reached, and that there will be no data anymore)

Programming Logic & Design Comprehensive

9th Edition

ISBN:9781337669405

Author:FARRELL

Publisher:FARRELL

Chapter7: File Handling And Applications

Section: Chapter Questions

Problem 15RQ

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What will be the content of 'file.c' after executing the following program? #include<stdio.h> int main() { FILE *fp1, *fp2; fp1=fopen("file.c", "w"); fp2=fopen("file.c", "w"); fputc('A', fp1); fputc('B', fp2); fclose(fp1); fclose(fp2); return 0; }
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Assume that infile has been properly defined to read data from a file and that a file of integers has been properly opened. If you don't know how many integers are in the file, which loop(s) would be appropriate to read in all the values from the file? Variable n has been defined as an int. 1. while(! infile.eof()) 2. for(int x = 0; x < 1000000; x++) 3. while(infile >> n) Answers: A. Both 2 and 3 B. 1, 2, and 3 C. Both 1 and 2 D. Both 1 and 3 E. 3 only
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Write a program which reads the content of two files “text1.txt” and “text2.txt” line byline and merges them into another file called “merge12.txt”.You can safely assume that the input is valid. (please write the code in programming C)
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I have the following code but it is not running correctly! import csvlst = []def save_dict_to_csv(storage, file_name): with open(file_name, 'w', newline = '') as f: f.write(file_name) f.write('\n') for key, value in storage.items(): lst1 = [] lst1.append(str(key)) for val in value : if type(val) == list: for v in val: lst1.append(str(v)) else: lst1.append(str(val)) for i in range(len(lst1)): if i != (len(lst1) - 1): f.writelines(lst1[i] + ',') else: f.writelines(lst1[i]) f.writelines('\n') f = open(file_name) print(f.read()) if __name__ == "__main__": storage = { 101: ["quiz", 12, [1,2,3,4]], 102: ["final", 53, [10, 10, 10, 10]] } res = save_dict_to_csv(storage,…
I have this code but , doesn't print on the file, so can you figure the error the code is #include <stdio.h> #include <stdlib.h> void charges (int usage , double *cost,double *avg); double round (double money); int main() { int id,month, year; double usage, cost , avg; FILE *fin ,*fout; fin = fopen("usage.txt","r"); fout= fopen("charge.txt","w"); fscanf(fin,"%d,%d",&month,&year); double valid = fscanf (fin,"%d,%lf",&id,&usage); while(valid != EOF) { valid = fscanf (fin,"%d,%lf",&id,&usage); charges (usage,&cost,&avg); double roundcost=round(cost); double roundavg=round (avg); fprintf(fout,"charges for %d\\%d",&month,&year); fprintf(fout,"%d %f %0.2f %0.2f",id,usage,cost,avg); } fclose(fin); fclose(fout); return 0; } void charges (int usage , double *cost,double *avg) { double money=0; if (usage <= 10) { *cost=7.99; } else { *cost =…
Task 1. Create a C file and declare declare the following variables in your main function: char string1[12] = "hello"; char string2[12] = "hello\n"; char string3[12] = "hello\n\0you\n"; char string4[12] = "0123456789AB"; The first function that we will write is the void printArray(char a[], int length) function. It has an array of characters as input and an integer that indicates the length of the array. void signifies that it has no return value. The function prints the '[' character and then all characters in the array, separated by commas, finally followed by the ']' and '\n' characters. When a character in the string is null then you should print \0, i.e. the backslash character followed by ’0’. When a character in the string is the newline character then you should print \n, i.e. the backslash character followed by ’n’. Call printArray with each of the string variables as input. The output should be as follows: [h,e,l,l,o,\0,\0,\0,\0,\0,\0,\0] [h,e,l,l,o,\n,\0,\0,\0,\0,\0,\0]…