Question

int main()
{
int a = 0;
int BUFSIZE = 1000;
char *string1[20];
FILE *fp1 = fopen("input1.txt", "r");
if (fp1 == 0)
{
    fprintf(stderr, "Error while opening");
    return 0;
}
string1[a] = (char *)malloc(BUFSIZE);
while (fgets(string1[a], BUFSIZE, fp1)!=NULL)
{
    a++;
    string1[a] = (char *)malloc(BUFSIZE);
} 
printf("%c", string1[3]);
}

Hi, I get the above code, which reads a string from a text file and store it in an array. Now I want to output a certain element of array string1, but apparently printf doesn’t work. Besides, what does char *string1[20] exactly define? Does it have something to do with pointer? Thank you！

Answer 1

char *string1[20]

declares an array, named string1, of 20 pointers to char. Thus string1[3] is a pointer to char, and not a char, as would be required for the %c format.

Since string1[3] was – if at all – filled via fgets, it contains a 0-terminated string, so you can print it out using

printf("%s\n", string1[3]);

If you want to print a single character, you’d use

printf("%c\n", string1[3][4]);

for example.