Pointer with strcat() and assignment

[waqaradil89]

Hello guys. I am stuck help me out.
I am trying to write a code that simply adds "s" in the end of name. In other words; Singular to plural (no ies es rule; just "s" in the end)
Code is here:

#include<stdio.h>
#include<conio.h>
#include<string.h>
main()
{
     char *ptrname[35];
     ptrname[0]="Cow";
     ptrname[1]="Parrot";
     for(int i=0;i<2;i++)
     {
     strcat(ptrname[i],"s");
     printf("%s\n",ptrname[i]);
     }
    
     getch();
}

[laserlight]

Use a 2D array instead, making sure that there's enough space in each inner array to hold both the 's' to be added and the null character. You will also need to initialise this 2D array instead of using assignment, or use strcpy.

At the moment, your pointers point to the first character of string literals, so you end up trying to modify string literals, resulting in undefined behaviour.

[rstanley]

Hello guys. I am stuck help me out.
I am trying to write a code that simply adds "s" in the end of name. In other words; Singular to plural (no ies es rule; just "s" in the end)
Code is here:

#include<stdio.h>
#include<conio.h>
#include<string.h>
main()
{
     char *ptrname[35];
     ptrname[0]="Cow";
     ptrname[1]="Parrot";
     for(int i=0;i<2;i++)
     {
     strcat(ptrname[i],"s");
     printf("%s\n",ptrname[i]);
     }
    
     getch();
}

ptrname[i] contains a pointer to a Constant String. You cannot strcat() a string onto a string constant.
main() should be int main(void), and you should return 0; at the end of your main function to indicate to the O/S that the app executed correctly, or a non-zero value if the app failed.
Please don't use conio, and it is non-standard.
I also recommend defining "int i" at the start of the main() function, not in the for() loop. Not allowed in all versions of the C standard.

[waqaradil89]

ptrname[i] contains a pointer to a Constant String. You cannot strcat() a string onto a string constant.

What is the solution ?

main() should be int main(void), and you should return 0; at the end of your main function to indicate to the O/S that the app executed correctly, or a non-zero value if the app failed.
Please don't use conio, and it is non-standard.
I also recommend defining "int i" at the start of the main() function, not in the for() loop. Not allowed in all versions of the C standard.

Thanks for the suggestion.

[waqaradil89]

You will also need to initialise this 2D array instead of using assignment, or use strcpy.

What does it mean? Is there any other way of initialisation?

[laserlight]

What does it mean? Is there any other way of initialisation?

In other words, if you write:

char ptrname[35][20];

then you cannot continue to write:

ptrname[0] = "Cow";
ptrname[1] = "Parrot";

Rather, you either have to use initialisation:

char ptrname[35][20] = {"Cow", "Parrot"};

or use strcpy or some other suitable function:

strcpy(ptrname[0], "Cow");
strcpy(ptrname[1], "Parrot");

[waqaradil89]

In other words, if you write:

char ptrname[35][20];

then you cannot continue to write:

ptrname[0] = "Cow";
ptrname[1] = "Parrot";

Rather, you either have to use initialisation:

char ptrname[35][20] = {"Cow", "Parrot"};

or use strcpy or some other suitable function:

strcpy(ptrname[0], "Cow");
strcpy(ptrname[1], "Parrot");

I got it. Thanks i will try your way.

[waqaradil89]

In other words, if you write:

char ptrname[35][20];

then you cannot continue to write:

ptrname[0] = "Cow";
ptrname[1] = "Parrot";

I think you are mistaken, i am asking for the pointer array.. you r suggesting variable array.. ptrname[][]... what i want is *ptrname[][]; and i tried this 2d aray with pointer , no success

[Salem]

It's not that you can't use pointers, it's just that the things you make it point to (things like "cow") are CONSTANTS.

You need to make them point to writeable memory,

     char *ptrname[35];
     ptrname[0] = malloc(10);
     strcpy(ptrname[0],"Cow");  // now it can be modified.
     strcat(ptrname[0],"s");

[rstanley]

I think you are mistaken, i am asking for the pointer array.. you r suggesting variable array.. ptrname[][]... what i want is *ptrname[][]; and i tried this 2d aray with pointer , no success

char ptrname[35][20];

This is fine. You are defining an array of 35, 20 character arrays that can hold up to 19 characters per array, plus one byte for the '\0' nul bytes.

The problem is with the code:

ptrname[0] = "Cow";
ptrname[1] = "Parrot";

You cannot copy strings through an assignment! You must use strcpy() to copy the strings:

strcpy(ptrname[0], "Cow");
strcpy(ptrname[1], "Parrot");

As @laserlight explained to you.

[grumpy]

I think you are mistaken, i am asking for the pointer array..

It is your usage that is mistaken.

For your manner of using them, your pointers in the array need to be initialised so they point to (the first element of) modifiable arrays.

    char *ptrname[35];
    char modifiable_buffer[35][20];
    for (i = 0; i < 35; ++i)
        ptrname[i] = modifiable_buffer[i];
    strcpy(ptrname[0], "Cow");        /*  this will copy characters from "Cow" into modifiable_buffer[0]   */
    strcpy(ptrname[1], "Parrot");
      /*  etc */

It is also possible to use malloc() to initialise each element of ptrname.

The point - which others have told you repeatedly - is that modifying something that is constant gives undefined behaviour. String literals like "Cow" are represented as const arrays. Modifying individual characters (e.g. change the 'C' to a 'c') or writing past the end (appending characters as in your example) each cause undefined behaviour. So the technique is to create a copy that can be modified, and then modify the copy.


You happen to be picking at an anomaly in the C language. By rights, the assignment "ptrname[0] = "Cow"" should be illegal, since it discards const'ness (it allows creating a non-const pointer to something that is const). The reasons for that are tedious historical concerns but doesn't change the basic premise: if your code uses that feature, your code is broken. However, if you turn up warning levels from your compiler, it should produce a warning about that.

[waqaradil89]

It's not that you can't use pointers, it's just that the things you make it point to (things like "cow") are CONSTANTS.

You need to make them point to writeable memory,

     char *ptrname[35];
     ptrname[0] = malloc(10);
     strcpy(ptrname[0],"Cow");  // now it can be modified.
     strcat(ptrname[0],"s");

I tried allocating memory but it returns error
[Error] invalid conversion from 'void*' to 'char*' [-fpermissive]

[laserlight]

Invalid assignment. malloc is a void and it returns nothing where as ptrname[] is a char. So it is not possible to assign a void to char. please Check again.

The return type of malloc is void* (i.e., pointer to void), not void. The type of ptrname[0] is char* (i.e., pointer to char), not char. A void* is convertible to char*, so there is no problem there.

[waqaradil89]

@grumpy Thank you for the suggestions. Now i have enabled WARNINGS of C standards. I am a newbie.

[waqaradil89]

The return type of malloc is void* (i.e., pointer to void), not void. The type of ptrname[0] is char* (i.e., pointer to char), not char. A void* is convertible to char*, so there is no problem there.

HOW?