Trying to create a strcat in c-CodePudding

So, I'm trying to code a strcat function using pointers, just for studying purposes.

#include <stdio.h>
#include <string.h>

char *strcpyy(char *dest, char *orig){
    char *tmp = dest;
    while (*dest   = *orig  );
    return tmp;
}

char *strcatt(char *dest, char *orig){
    strcpyy(dest   strlen(dest), orig);
    return dest;
}



int main(){
    char *a = "one";
    char *b = "two";
    printf("%s", strcatt(a,b));
}

When I run this code, the output is empty. Can anyone point out the problem?

CodePudding user response：

String literals are read-only. Any attempt to write to a string literal will invoke undefined behavior, which means that your program may crash or not behave as intended.

Therefore, you should not use a pointer to a string literal as the first argument to strcat or your equivalent function. Instead, you must provide a pointer to an object which is writable and has sufficient space for the result (including the terminating null character), such as a char array of length 7. This array can be initialized using a string literal.

Therefore, I recommend that you change the line

char *a = "one";

to the following:

char a[7] = "one";

After making this change, your program should work as intended.

CodePudding user response：

You declared two pointers to string literals

char *a = "one";
char *b = "two";

You may not append one string literal to another.

Instead you need to define the variable a as a character array large enough to contain the appended string literal pointed to by the pointer b.

And the both functions should be declared like

char *strcpyy(char *dest, const char *orig);

char *strcatt(char *dest, const char *orig);

Also as you are using standard C string functions like strlen

strcpyy(dest   strlen(dest), orig);

then it will be logically consistent to use standard C function strcpy instead of your own function strcpyy.

Otherwise without using standard string functions your function strcatt can look the following way

char * strcatt( char *s1, const char *s2 )
{
    char *p = s1;

    while ( *p )   p;
    while ( ( *p   = *s2   ) != '\0' );

    return s1;
}

Here is a demonstration program.

#include <stdio.h>

char * strcatt( char *s1, const char *s2 )
{
    char *p = s1;

    while ( *p )   p;
    while ( ( *p   = *s2   ) != '\0' );

    return s1;
}

int main( void ) 
{
    char a[7] = "one";
    const char *b = "two";

    puts( strcatt( a, b ) );
}

The program output is

onetwo

CodePudding user response：

You cannot modify "string literals". Those are not mutable.

The usual idiom for this sort of operation is to build up a string in a temporary working buffer that should be pre-dimensioned large enough to hold all that is required.

The following also shows more obvious code in both your functions.

#include <stdio.h>

char *strcpyy( char *dst, const char *org ) {
    for( char *p = dst; (*p   = *org  ) != '\0'; /**/ )
        ; // loop
    return dst;
}

char *strcatt( char *dst, const char *org ) {
    char *p = dst;
    while( *p != '\0' )
        p  ; //loop
    while( (*p = *org  ) != '\0' )
        p  ; // loop
    return dst;
}

int main(){
    const char *a = "one ";
    const char *b = "two ";
    const char *c = "three";

    char wrk[ 64 ]; // sufficient mutable space defined

    printf( "%s\n", strcatt( strcatt( strcpyy( wrk, a ), b ), c ) );

    return 0;
}

one two three