Caller cannot see the char array string returned from a function

Question

I am returning a char pointer from a function. But the caller is unable to see the string.

char* getFileContent(char* file)
{

FILE* fp = fopen("console.txt", "r");

fseek(fp, 0L, SEEK_END);
size_t sz = ftell(fp);
fseek(fp, 0L, SEEK_SET);


char* message = (char*)malloc(sz+1);
char buf[sz+1];

size_t len = 0;
if (fp != NULL)
{
    len = fread(buf, sizeof(char), sz, fp);
}

printf("MALLOC SIZE:%d FILE SIZE:%d", sz, len);

strcpy(message,buf);    //Modified code. This line fixed the code
message[++len] = '\0';
//printf("MESSAGE:%s", message);

return(message);

}

This is the caller. Output is empty.

int main(int argc, char **argv, char **env)
{

char* msg = getFileContent(imagefile);

if(msg != NULL)
    printf("Output:%s \n", msg);

free(msg);

return 0;
}

Please help.

What is the output of your program?

rob mayoff
– rob mayoff

2011-11-26 06:41:14 +00:00
Commented Nov 26, 2011 at 6:41 — rob mayoff
– rob mayoff, Commented Nov 26, 2011 at 6:41

Mysticial · Accepted Answer · 2011-11-26 06:51:54Z

1

The error is here:

printf("Output:", msg);

You're not printing the string. Try this instead:

printf("Output: %s", msg);

The %s is needed to tell printf() to print msg as a string.

Note that due to buffering, you may also need to add a \n:

printf("Output: %s \n", msg);

Here's another minor error:

message[++len] = '\0';

should be:

message[len] = '\0';

edited Nov 26, 2011 at 6:51

answered Nov 26, 2011 at 6:40

Mysticial

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14 Comments

cppcoder Over a year ago

Still output is empty. But, if I do strcpy in the function, it is coming in the output. But file contents are not.

Mysticial Over a year ago

Did you try adding the \n into the printf(). I edited this into my answer a minute later, so you may have missed it. Other than that, I don't see anything else wrong with the code - unless the file has a null character somewhere in the beginning.

cppcoder Over a year ago

If I modify the code like this, it works.

char* message = (char*)malloc(sz+1);     char buf[sz+1];      size_t len = 0;     if (fp != NULL)     {         len = fread(buf, sizeof(char), sz, fp);     }      printf("MALLOC SIZE:%d FILE SIZE:%d", sz, len);     strcpy(message,buf);

Brian Campbell Over a year ago

@cppcoder Can you edit your question to include the exact current code in your program (including the changes suggested by Mysticial), and the exact output that you get?

Mysticial Over a year ago

One more small mistake I found: message[++len] = '\0'; should be message[len] = '\0'; I'll add this to my answer.

|

LittleFunnyMan · Accepted Answer · 2011-11-27 09:27:03Z

strcpy(message,buf);    //Modified code. This line fixed the code
message[++len] = '\0';

I'm not sure if it is guaranteed that after the allocation of the buffer buf it is nullified. Therefore buf[len] might not be '\0' after reading and if there is no '0\' in the read text then your strcpy might run outside its bounds. So either use strncpy or change the above two lines into

buf[len++] = '\0\;
strcpy(message,buf);

Also the first version places the '\0' at len+1 which in my opinion is false. Imagine you've read 0 bytes then len=0 and message[1] is set to '\0' which leaves message[0] undefined.

Probably your code just ran fine because there is a create change the the allocated buffer buf is either filled with zeros or your compiler actively nullifies it. But as far as I know this is not mandatory for C compilers todo.

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