C - Error in Comparing Variable of Struct and String

Question

I am trying to implement a function that compares two strings. However, i am getting "expression must be a modifiable lvalue" in function strcmp even if i use char pointers instead of char array. My code is in following.

typedef enum { false, true } boolean;

struct threeGram *threeGram_array;

struct threeGram
{
    const char *value;
    int occurence;
};

boolean containsValue(struct threeGram array[], const char *string){

    int i;
    for(i = 0; i < sizeof (array) / sizeof (struct threeGram); i++){
        if(strcmp(array[i].value, string) = 0){
            return true;
        }
    }
    return false;
}

And unrelated, but `sizeof (array) / sizeof (struct threeGram)` does not give you the size of the array, since `array` decayed to a pointer — UnholySheep, Oct 04 '18 at 14:34
Do you suggest any thing for to get max limit of loop ? I am trying to get the length of my array passed into function... — Selim Nuhoğlu, Oct 04 '18 at 14:38
Get a better compiler. You should get a warning/error for this code. The result of strcmp is not even a lvalue. Out of curiousity, which trash compiler let this code through? — Lundin, Oct 04 '18 at 14:40
@Lundin OP is already getting a compiler **error**. In what sense is a warning better? — Ivaylo Strandjev, Oct 04 '18 at 14:42
@Lundin I am using Visual Studio Code. Which one should i use ? Can you suggest one ? — Selim Nuhoğlu, Oct 04 '18 at 14:42

score 1 · Accepted Answer · answered Oct 04 '18 at 14:35

1

When checking for equality in C you should use double equals == otherwise this is interpreted as assignment. Thus on this line:

if(strcmp(array[i].value, string) = 0){

You try to assign a result to function return value which is non-modifiable lvalue.

answered Oct 04 '18 at 14:35

Ivaylo Strandjev

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C - Error in Comparing Variable of Struct and String

1 Answers1