Great god help me to look at the trouble running result is what is wrong-CodePudding

? (T? T)? I stuck two hours the meowed

The first figure is the title, the second figure is my run results

#include
Int main ()
{
Int a [3], [3], I, t, j;
Printf (" do put in: \ n ");
for(i=0; i<3; I++)
{
for(j=0; j<3; J++)
{
The scanf (" % d ", & amp; A [I] [j]);
If (a [I] [j] % 3!=0)
Printf (" % 5 d, "a [I] [j]);
The else
Printf (" % 5 d \ n ", a [I] [j]);

}

}

T +=a, [I] [j].
Printf (" The result is=% d \ n ", t);
}

CodePudding user response:

T uninitialized began to t=0

CodePudding user response:

I didn't learn c + +!!!!! But I see ~ ~ t +=a, [I] [j]. This should be a cumulative process!!! The additive process outside loop???? Don't know how to calculate your t is the result?

CodePudding user response:

T didn't initialize, define variables when assigning a 0, and the t +=a [I] [j] why didn't in the for loop?

CodePudding user response:

#include
Int main ()
{
Int a [3], [3], I, j;
Int t=0;
Printf (" do input: \ n ");
for(i=0; i<3; I++)
{
for(j=0; j<3; J++)

{
The scanf (" % d ", & amp; A [I] [j]);
If (j==2)
Printf (" % 5 d \ n ", a [I] [j]);
The else
Printf (" % 5 d, "a [I] [j]);
T +=a, [I] [j].
}

}

Printf (" The result is: % d \ n ", t);
}

CodePudding user response:

refer to the second floor limeilu110 response:

I didn't learn c + +!! But I see ~ ~ t +=a, [I] [j]. This should be a cumulative process!!! The additive process outside loop???? Don't know how to calculate your t is the result?

Maybe because I didn't give t assignment it literally find number?

CodePudding user response:

Reference:

 # include 
Int main (int arg c, char * argv []) 
{
Int a [3], [3], I, t=0, j. 
Printf (" do put in: \ n "); 
for(i=0; i<3; I++) 
for(j=0; j<3; J++) 
The scanf (" % d ", & amp; A [I] [j]); 

for(i=0; i<3; I++) 
for(j=0; j<3; J++) 
{
Printf (" % 5 d, "a [I] [j]); 
If ((j + 1) % 3==0) 
printf("\n"); 
} 

for(i=0; i<3; I++) 
for(j=0; j<3; J++) 
{
T +=a, [I] [j]. 
} 
Printf (" The result is=% d \ n ", t); 

return 0; 
}

CodePudding user response:

Simplify the step:

 # include 
Int main (int arg c, char * argv []) 
{
Int a [3], [3], I, t=0, j. 
Printf (" do put in: \ n "); 
for(i=0; i<3; I++) 
for(j=0; j<3; J++) 
{
The scanf (" % d ", & amp; A [I] [j]); 
} 

for(i=0; i<3; I++) 
for(j=0; j<3; J++) 
{
T +=a, [I] [j]. 
Printf (" % 5 d, "a [I] [j]); 
If ((j + 1) % 3==0) 
printf("\n"); 
} 
Printf (" The result is=% d \ n ", t); 

return 0; 
}

To simplify the two step, input methods:

 # include 
Int main (int arg c, char * argv []) 
{
Int a [3], [3], I, t=0, j. 
Printf (" do put in: \ n "); 
for(i=0; i<3; I++) 
for(j=0; j<3; J++) 
{
The scanf (" % d ", & amp; A [I] [j]); 
Printf (" % 5 d, "a [I] [j]); 
If ((j + 1) % 3==0) 
printf("\n"); 
T +=a, [I] [j]. 
} 
Printf (" The result is=% d \ n ", t); 

return 0; 
}

CodePudding user response:

This is running out for you, but I didn't understand in the beginning the pointer? An array is doing

CodePudding user response:

The

reference

int main (int arg c, char * argv [])

do you speak is it? This is just a compiler is different, the format of the generated different, you can put the deleted in brackets

 int main ()

CodePudding user response:

references 9 f QZJHJXJ response:

reference
int main (int arg c, char * argv [])

do you speak is it? This is just a compiler is different, the format of the generated different, you can put the deleted in brackets
int main ()

Thank you, I understand.