I'm writing a simple program in c, that prints all elements of an array. The array is

char op[2][50] = {"option1", "option2"};

the program is

int main(int argc, char * argv[])
{
    char spaces[6] = "      ";
    int choice;
    char op[2][50] = {"option1", "option2"};

    printf("\n\n\t=========== OPTIONS ===========\n\n\t");
    for (int i;i<sizeof(op)/sizeof(op[0]);i  ) {printf("%s[%d]: %s\n\t",spaces,i 1,op[i]);}
    printf("\n\t\t%s[?]: ",spaces);
    scanf("%d",&choice);
    
    if (choice==1) {firstFunction();} //i've declared this function, but here isn't important

    return 0;
}

So, the problem is that the options arent printed. output:

=========== VIDEO TOOLS ===========


               [?]:

the problem is essentially that if statement, because i've tried to add other printf up and down it and them works. Also, that if statement worked before some changes to the program, so the problem can be other lines of code(?)

i'm a beginner with c, so please not expose complex solutions. thanks

CodePudding user response：

Lots of bugs. Summary of things mentioned in comments:

• i is used while uninitialized, set it to zero.

• char spaces[6] = " "; is too small, see How should character arrays be used as strings?.

• Make it a habit to end your printf lines with \n rather than to begin with them. Because \n does not only change the line, it may also flush the stdout on some systems. Failing to flush stdout might cause the output to appear in a strange order or go missing.

Also in case these are read-only strings, then consider using a 1D array of pointers instead, since that's more flexible and memory efficient:

const char* op[2] = {"option1", "option2"};

CodePudding user response：

This array declaration where the number of elements is specified explicitly

char spaces[6] = "      ";

is error prone. The array does not contain a string because the terminating zero character '\0' of the string literal used as an initializer is not stored in the array spaces. The size of the string literal is equal to 7 not to 6.

So it is more safer to write

char spaces[] = "      ";

Or as you are not going to change the array then it will be better to write

const char *spaces = "      ";

In this declaration of the two-dimensional array

char op[2][50] = {"option1", "option2"};

you explicitly specified that the array contains two elements. So using the expression i<sizeof(op)/sizeof(op[0]) does not make a great sense.

Either introduce a named constant be fefor teh declaration of teh array like for example

enum { N = 2 };
char op[N][50] = {"option1", "option2"};

and then use the expression

i < N

in the for loop. Or declare the array like

char op[][50] = {"option1", "option2"};

and after the declaration introduce a named constant like

const size_t N = sizeof( op )/sizeof( op[0] );

and use the constant in the for loop where by the way you forgot to initialize the variable i

for ( size_t i = 0; i < N; i   ) 
{
    printf("%s[%zu]: %s\n\t",spaces,i 1,op[i]);
}

As the variable i now has the type size_t then you have to use the conversion specifier %zu instead of %d in the call of printf.

It will be safer to rewrite this code snippet

scanf("%d",&choice);

if (choice==1) {firstFunction();} //i've declared this function, but here isn't important

the following way

if ( scanf("%d",&choice) == 1 && choice==1 ) 
{
    firstFunction();
} //i've declared this function, but here isn't important

CodePudding user response：

for (int i;i<sizeof(op)/sizeof(op[0]);i )

You do not initialize the i variable and it is an Undefined Behaviour (UB)

It should be:

for (int i = 0;i<sizeof(op)/sizeof(op[0]);i  )

or better (using the correct type for sizes and positive indexes)

for (size_t i = 0;i<sizeof(op)/sizeof(op[0]);i  )

To print size_t use %zu format

Also, remember that C string is terminated by the null terminating character. So you need to add one more character than seen in the string literal. ("123" requires 4 characters) spaces array is too short to accommodate " ". char spaces[] = " ";