如何在Linux终端上用C打印此代码中的所有字符？

Question

I want to read a file and print a line from that file. Here is the code.

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

char* get_next_line(FILE* fpntr);

int main()
{
    FILE* fp = fopen("movies.txt", "r");
    char* tmp = get_next_line(fp);
    printf("%s", tmp);
    fclose(fp);
    return 0;
}

char* get_next_line(FILE* fpntr)
{
    char buff[2048];
    int index = 0;
    int ch = fgetc(fpntr);
    while(ch != '\n' && ch != EOF)
    {
        buff[index++] = ch;
        ch = fgetc(fpntr);
    }
    buff[index] = '\0';
    char* tmp;
    tmp = (char*)malloc((int)(index)*sizeof(char));
    strcpy(tmp,buff);
    return tmp;
}

Here is the output as displayed from Ubuntu terminal.

The first line in my movies.txt file is 1.Lord of the rings the fellowship of the ring , but only the last few characters are being printed out.So I need help to print out the entire line instead of just the last few characters.

Answer 1

Your program prints a line without a line terminator. Apparently, your shell prompt contains code to return the cursor to the left margin before printing the prompt text; and so, the prompt replaces part of the program's output.

With a simpler prompt, you would get something like

bash$ ./new
1. Lord of the rings the fellowship of the ringbash$

where bash$ is your prompt.

If this is not what you want, printf("%s\\n", ...) would be the normal and expected way to print a line with a newline added; or, you could avoid trimming the newline in the first place. If the program isn't one you can change yourself, adding a newline after running it could be done with

bash$ ./new; echo

If you change your prompt to always print an empty line before the prompt text, you can avoid this problem altogether, but then you will typically want a rather big terminal window. (I see you already have one, but I guess that's just because you tolerate Ubuntu's insane default settings.)