How to compare a string from an array in C?

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How to compare a string from an array in C?

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#1 by (3 votes)

1

I have the following variables:

char nome[10][8];
int i; --> Sendo esta para o loop

I ask the user to enter 5 names and then I read the variable:

printf("Entre 5 nomes : \n");
for(i=0; i< 5 ; i++)
{
scanf("%s",nome[i]);
}

But after this, I want to compare a string from the array and see if the name is correct, and if it is, it triggers a message.

My attempt:

for (i=0; i < 1; i++) {
if (nome[i] == "Lucas de Oliveira");
printf("Idade : 18 anos, \nSexo : Masculino, \nEstado civil : Solteiro");
continue;
}

It appears the message, but it does not even finish the first for to fill in all the names, and the check is somewhat incorrect.

Screenshot:

How can I proceed to make the check correct, and if only the name that is entered that is within if , the message appears.

c if for

asked by anonymous 29.03.2017 / 07:55

1 answer

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score 3 · Accepted Answer

Problems:

You are using the wrong function to read the string.

You are not comparing the string, but the address it is in.

Little space in array.

It is not iterating as often as needed.

Start by arranging the array size:

char nome[5][20];
int i;

In an array like this, what you have are 5 lines and 20 columns , and the Marcos de Oliveira name has at least 18 characters , which should be ideal.

puts("Entre 5 nomes :");
for(i=0; i< 5 ; i++)
{
   scanf("%19[^\n]%*c",nome[i]);
}

In this part you can not use printf with \n , puts is much simpler. Of course% wont stop reading when finding empty space when used in conjunction with scanf , using this expression %s causes empty spaces to be used.

for(i=0; i < 5; i++) {
    if(!(strcmp(nome[i], "Lucas de Oliveira"))){
        printf("\nNome: %s\nIdade : 18 anos, \nSexo : Masculino, \nEstado civil : Solteiro", nome[i]);
        break;
    }
}

Here, you just need to change the expression in the %19[^\n]%*c arguments using the if function that compares two expressions, and returns strcmp if they are identical, and also 0 i < 1 that is the size of the array, otherwise it stops at the first iteration, and it does not advance to the remaining names, and of course, remove the i < 5 and put continue in place, so the iteration stops when finding the match.

Using fgets

As an alternative to expression within break , you could use scanf :

for(i=0; i< 5 ; i++)
{
    fgets(nome[i], 26, stdin);
}

As fgets keeps the fgets break, you must remove before you can compare with the desired word.

for(i=0; i < 5; i++) {
        if ((strlen(nome[i])>0) && (nome[i][strlen(nome) - 1] == '\n')){
            nome[i][strlen(nome) - 1] = 'char nome[5][20];
int i;
';
        }   
        if(!(strcmp(nome[i], "Lucas de Oliveira"))){
            printf("\nNome: %s\nIdade : 18 anos, \nSexo : Masculino, \nEstado civil : Solteiro", nome[i]);
            break;
        }
    }

If none or any of the functions used here are not working, see the documentation for another alternative, or check the compiler settings in use.