char and pointer in C language

Question

I have a method that returns me a pointer to characters, in the main file when I print I have the complete string, when I look at the length I only have 8 (up to character "c") when I loop over it I only have 8 characters. why when i print "p" i have my whole string and when i loop i only have 8 characters ? can you help me understand because I would like to stock all the characters to process them ?

method :

char* getData(char* file) {
    FILE* f = fopen(file, "r");
    char *string = malloc(250);
    int i = 0;
    if(f) {
        int c;
        printf("\nopen file to get data : %s\n", file);
        while((c = getc(f)) != EOF) {
            string[i] = c;
            i++;
        }
        string[i] = '\0';
        fclose(f);
    } else {
        printf("\nopen file to get data error...\n");
    }
    return string;
};

main file :

char* file = "target/to/message.txt";
char* p;

p = getData(file);
printf("\npointer in main file :\n--%s--\n", p);
printf("\n----------------\n");

printf("\nlen pointer in main file : --%llu--\n",sizeof(p));
printf("\n----------------\n");

for(unsigned long long i=0;i<sizeof(p);i++) {
    printf("\nloop to pointer p : --%c--",p[i]);
}

out of console :

Mareek Roy · Accepted Answer · 2022-08-11 12:00:59Z

1

use strlen() instean of sizeof for char* variables, sizeof() is giving you the address of char* p which is 8 byte long and thus only few characters are getting printed.

edited Aug 11, 2022 at 12:00

answered Aug 11, 2022 at 7:21

Mareek Roy

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1 Comment

mwck46 Over a year ago

Actually sizeof(pointer) is 8 byte long on a 64-bit machine and 4 byte long on a 32-bit machine.

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