How to insert a ";" at the end of the string, program in c

Question

I wonder what I should do for the ";" in the print appear afterward of string and not before. I’m new to programming and I’d like to appeal to your help.

Thank you

Input

Linha1
Linha2
Linha3
Linha4

Expected output

Produtos:
- Linha3;
- Linha4;
- Linha2;
- Linha1;

Output obtained

Produtos:
- Linha1
;- Linha2
;- Linha3
;- Linha4
;

Code

#include <stdio.h>
#include <string.h>
#include <stdlib.h>   

typedef struct linha {
  char *produtos;
  struct linha *prev;
} linha;

FILE *file;   
int main() {

  linha *tail = NULL;
  file = fopen("text.txt","r");
  char linha1[255];

  while (fgets(linha1,255,file)!=NULL) {
    linha *l1 =(linha*) malloc(sizeof(linha));
    l1->produtos =(char*) malloc(strlen(linha1)+1);
    strcpy(l1->produtos,linha1);
    l1->prev = tail;
    tail = l1;
  }

  linha *current;
  current = tail;
    printf("Produtos:\n");
  while (current != NULL) {
    printf("- %s;",current->produtos);
    current = current->prev;
  }
  return 0;

}

Answer 1

Edited, because if the file being read did not end with a line break ('\n'), the last character would be deleted. A test is placed to verify that the last character of the line read is '\n' before replacing.

As the user72726 explained in his reply, the fgets function includes line advance when reading from the file ('\n'). At first I thought of inserting the semicolon when reading the file, but this limits some other treatment you may want to do.

I suggest taking this line forward while reading from the archive

while (fgets(linha1,255,file)!=NULL) {
  linha *l1 =(linha*) malloc(sizeof(linha));
  if (linha1[strlen(linha1) - 1] == '\n') // testa se o último caractere é \n
    linha1[strlen(linha1) - 1] = '\0'; // troca o \n pelo char null
  l1->produtos =(char*) malloc(strlen(linha1)+1);
  strcpy(l1->produtos,linha1);
  l1->prev = tail;
  tail = l1;
}

and then add the semicolon and the line advance when displaying on the screen

while (current != NULL) {
  printf("- %s;\n",current->produtos);
  current = current->prev;
}

It works because functions that handle strings stop when finding a null character - see the ref. from the following link strcpy, at the end of the first sentence "including the terminating null Character (and Stopping at that point)".

Note that if done sizeof(linha1) will look different from strlen(linha1), the normal, if I’m not mistaken, is the difference being 1 but here will be 2 because linha1 is a pointer that at the end has two null characters instead of 1, which strlen does not count.

The above paragraph is incorrect because sizeof(linha1) in the case of this program will always be 255, as it is the size allocated in the declaration.

One important thing, though off the subject of the question, is to release all the memory you use. So, if it were me, for each malloc I would put a memfree on the program even if the program was to close soon after.

References (other functions appear in the menu on the left): http://www.cplusplus.com/reference/cstring/strlen/ http://www.cplusplus.com/reference/cstring/strcpy/

Answer 2

The function fgets returns all line characters, including the end-of-line character itself - \n (obviously, it is the last character of the string returned). And that is why the printing is not being done in the desired way.

Then, you could choose to replace the last character of the string with a blank.

Ex:

linha1[strlen(linha1) -1 ] = ' ';
strcpy(l1->produtos,linha1);

Answer 3

The code entered in the question is not in C style++.

I rewrote using C++ libraries. Compiled with:

g++ -Std=c++11 stackoverflow.cpp -o stackoverflow

#include <iostream>
#include <fstream>

using namespace std;

int main() {

    ifstream entrada("entrada.txt");
    string linha;

    while (getline(entrada, linha)) {
        cout << " - " << linha << ";" << endl;
    }

    return 0;
}