bitman
asked on
newbie needs help on easy one
I am new to C and have written a simple program ( added below) that receives a string allocates address to a pointer
allocates memory to the string accordingly and is supposed to print the string. I think I did everythig right so why does it print garbage instead of the string ???
allocates memory to the string accordingly and is supposed to print the string. I think I did everythig right so why does it print garbage instead of the string ???
Post the code.
Where is the code?
Its possible that it hasn't been initialised with a NULL value at the last character, but would need the code to verify.
If this is the problem then the code could be something like
myarr[10] = '\0';
10 being the last value of the array.
If this is the problem then the code could be something like
myarr[10] = '\0';
10 being the last value of the array.
You need to allocate the memory before you put the string in it.
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Hi Deighton
I'm a bit confused. If we malloc before gets it would mean that we are mallocing the pointer itself. In the case of my program is it worth mallocing the pointer as from what I understand one pointer is a fixed 2 or 4 bytes. I can understand mallocing an array of pointers before gets but not a single pointer.
Also the string itself, surely this can only be malloced the gets, else how does the compiler know the strlen ?
I actually succeed
Please clarrify,
thanks for everything
Bitman
I'm a bit confused. If we malloc before gets it would mean that we are mallocing the pointer itself. In the case of my program is it worth mallocing the pointer as from what I understand one pointer is a fixed 2 or 4 bytes. I can understand mallocing an array of pointers before gets but not a single pointer.
Also the string itself, surely this can only be malloced the gets, else how does the compiler know the strlen ?
I actually succeed
Please clarrify,
thanks for everything
Bitman
0 points? Cool! I could do with zero points...
It is interesting that this answer (below) was accepted. Let's have a look at what's wrong with it:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main()
main returns int, not void. Any other return type, and you're no longer programming in C, but in some bizarre concoction of your own.
{
char *ptr,str[20];
ptr = str;
This assignment is fine, but pointless, as the value of ptr is about to be overwritten immediately.
tr = (char*)malloc(strlen(ptr)+
Firstly, at this point, str's contents are uninitialised, so the amount of memory allocated here is more or less random.
Secondly, why the cast? It's completely unnecessary. If your compiler gives you a diagnostic for it, you're not using a C compiler (in C++, the cast IS required, and some editors save as .cpp rather than .c if you are not firm with them).
if(ptr == NULL)
{
printf("memory allocation failure");
return;
main returns int. This should be return EXIT_FAILURE;
}
printf("Please enter a string >:");
If you want to /ensure/ that this output appears in time for the user, either add a \n to the end of the prompt, or call fflush(stdout);
gets(ptr);
gets() is a bug waiting to happen. Don't /ever/ use it. It doesn't protect its buffer against overflow.
Use fgets() instead. Note that, for fgets, you will need a decent buffer, not one that has a random size, such as in this program.
printf("%s\n\n",ptr);
If you fix everything else that's wrong with this program, this printf should work just fine. :-)
Once you're done with memory, it's a good plan to free it up. One day, this bit of defensive programming will save you a LOT of problems. (I speak from bitter experience here.)
free(ptr);
Because main returns int, you should add
return 0;
or
return EXIT_SUCCESS;
Either will do.
}
I am deeply concerned that you accepted this answer, as it had more bugs in it than an ant farm.
There is something amiss with a system which allows such answers to be accepted.
It is interesting that this answer (below) was accepted. Let's have a look at what's wrong with it:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main()
main returns int, not void. Any other return type, and you're no longer programming in C, but in some bizarre concoction of your own.
{
char *ptr,str[20];
ptr = str;
This assignment is fine, but pointless, as the value of ptr is about to be overwritten immediately.
tr = (char*)malloc(strlen(ptr)+
Firstly, at this point, str's contents are uninitialised, so the amount of memory allocated here is more or less random.
Secondly, why the cast? It's completely unnecessary. If your compiler gives you a diagnostic for it, you're not using a C compiler (in C++, the cast IS required, and some editors save as .cpp rather than .c if you are not firm with them).
if(ptr == NULL)
{
printf("memory allocation failure");
return;
main returns int. This should be return EXIT_FAILURE;
}
printf("Please enter a string >:");
If you want to /ensure/ that this output appears in time for the user, either add a \n to the end of the prompt, or call fflush(stdout);
gets(ptr);
gets() is a bug waiting to happen. Don't /ever/ use it. It doesn't protect its buffer against overflow.
Use fgets() instead. Note that, for fgets, you will need a decent buffer, not one that has a random size, such as in this program.
printf("%s\n\n",ptr);
If you fix everything else that's wrong with this program, this printf should work just fine. :-)
Once you're done with memory, it's a good plan to free it up. One day, this bit of defensive programming will save you a LOT of problems. (I speak from bitter experience here.)
free(ptr);
Because main returns int, you should add
return 0;
or
return EXIT_SUCCESS;
Either will do.
}
I am deeply concerned that you accepted this answer, as it had more bugs in it than an ant farm.
There is something amiss with a system which allows such answers to be accepted.
Add like a comment a simple program