如何正确打印交流字符串中的每个元素？

Question

#include <stdio.h>
#include <iostream>

using namespace std;

int main(){
    char *a[20];
    FILE * fin = fopen("testtest.txt","r");
    int i;

    fscanf(fin,"%s",a);

    for(i=0;i<20;i++)
    {
        printf("%c\n",a[i]);
    }

    system("pause");
}

In this program, I suppose to print each element in the array, which should be ABCDE but actually it prints:

It seems every element is weired, how should I print it correctly?

A
E
─
╒
┴
■
┌
·
Φ

8
↔


p
╘
╠
x
3
☻

Answer 1

The type of a is an array of char* , not an array of char . Change to:

char a[20];

Recommend compiling at the highest warning level and treat warnings as errors. For example:

$ gcc -Wall -Werror -pedantic main.c
main.c: In function ‘main’:
main.c:9:5: error: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char **’ [-Werror=format]
main.c:10:5: error: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Werror=format]
cc1: all warnings being treated as errors

Check the result of fopen() and fscanf() to be certain the file was opened and data was read into a before attempting to use the variables.

Answer 2

a is an array of 20 char pointers . I think you wanted an array of 20 chars instead

char a[20];

Answer 3

char a[20];
FILE * fin = fopen("testtest.txt","r");
int i;

fscanf(fin,"%19s",a);