lvalue required as left operand of assignment - What causes this error and how to fix it?-CodePudding

Here is the piece of code. Its in C, compiler is CodeBlocks I made a function Replacethings which is supposed to replace all characters that are spaces, commas, and exclamation points with *. Seemingly an easy task, and yet so hard. The input has to be 25 characters or less.

void Replacethings( char *StrongOfChars[25]){
 for(int i =0; i<25; i  ){

    if(*StrongOfChars[i]=' ' || *StrongOfChars[i] = ',' || *StrongOfChars[i] = '!'){
        *StrongOfChars[i]= '*';
        printf("%c", StrongOfChars[i]);
    }

 }
} 
printf("Enter your favorite quote (Has to be less than 25 characters)!");
    char StronKK[25];

    scanf("%s", &StronKK);
    Replacethings(&StronKK);
    printf("\n Your favorite squote is now: %s  ", StronKK);

CodePudding user response：

*StrongOfChars[i]=' ' means you assign the value ' ' to the first character of the i'th char pointer:

    if(*StrongOfChars[i]=' ' || *StrongOfChars[i] = ',' || *StrongOfChars[i] = '!'){

You want to use == instead. The first expression of the body of the if statement is:

*StrongOfChars[i]= '*';

which probably should be part of your if expression instead, i.e.:

if(
   *StrongOfChars[i] == ' ' ||
   *StrongOfChars[i] == ',' ||
   *StrongOfChars[i] == '!' ||
   *StrongOfChars[i] == '*'
) {
   ...
}

or you can use strchr(" ,!*", *StrongOfChars[i]) instead of those 4 conditions.

The code is poorly formatted, so you cannot see that the } before the 2nd print statement is in the wrong place so the print statement is not in a function. Maybe remove it or you forgot a main() function?

CodePudding user response：

As you declared the array StronKK like

char StronKK[25];

then the expression &StronKK has the type char ( * )[25].

So you have to write in the call of scanf

scanf( "$s", StronKK );

On the other hand, the function parameter has the type char *StrongOfChars[25]

void Replacethings( char *StrongOfChars[25])

that is adjusted by the compiler to the type char **

void Replacethings( char **StrongOfChars )

The types char ( * )[25] and char ** are not compatible types and the compiler should issue a diagnostic message.

You need to declare the function like

void Replacethings( char *StrongOfChars )

void Replacethings( char StrongOfChars[25])

void Replacethings( char StrongOfChars[])

and call it like

Replacethings( StronKK );

Pay attention to that the user can enter a string with the length less than 25. So using the magic number 25 within the function is a bad idea and can result in undefined behavior.

Also instead of the equality operator == you are using the assignment operator = within the for loop of the function.

The function can be defined the following way

void Replacethings( char StrongOfChars[])
{
    for ( ; *StrongOfChars != '\0';   StrongOfChars )
    {
        if ( *StrongOfChars == ' ' || 
             *StrongOfChars == ',' || 
             *StrongOfChars == '!' )
        {
            *StrongOfChars = '*';
        }
    }
}

It would be also reasonable to check whether a character is a tab character. That is the if statement can look

        if ( *StrongOfChars == ' ' || 
             *StrongOfChars == '\t' || 
             *StrongOfChars == ',' || 
             *StrongOfChars == '!' )
        {
            *StrongOfChars = '*';
        }

Or you can write it lime

if ( strchr( " \t,!", *StrongOfChars ) )
{
    *StrongOfChars = '*';
}

To use the function strchr you need to include header <string.h>.