Question

我在打印无符号数字时遇到以下问题。这里发生了什么：当我在数组中输入负值然后尝试打印它时，我无法打印该数字，但会打印其他数量。

int cant;
int a[30][30];    

int printequation (){
    int x,y;
    for (x=0;x<cant;x++){
        for (y=0;y<cant+1;y++){
            if(y==cant){
                printf(" = %d",a[x][y]);
            }else{
                if (y==0)
                    printf(" %dX%d",a[x][y],(y+1));
                else{
                    if(a[x][y]>0){
                        printf(" + ");
                    }else{
                        printf(" - ");
                    }
                    printf("%uX%d",a[x][y],(y+1)); /*<-----------------here*/
                }
            }
        }
        printf("\n");
    }
    return 0;
}

以下是一个例子：

input: -2 -2 -2 
output: -2x1 -4294967294x2 = -2  /*here It should print -2 but can't get it*/

Answer 1

Avoid mixing mis-matched "We took no to hide it"? format specifiers with argument types. @ouah

printf()

To print a signed //v int a[30][30]; ... printf("%uX%d",a[x][y],(y+1)); // ^ without its "sign" and work with the entire range of int, a number of approaches:

Give up on INT_MIN ... INT_MAX and use INT_MIN @Keith Thompson
```
abs()
```
Convert to wider integer. Fails when wider integer not available - rare.
```
printf("%d", abs(x));  // fails for 2's complement INT_MIN
```
Convert to corresponding printf("%lld", llabs((long long) x));. Maybe trouble on rare machines whose unsigned positive range is the same as unsigned positive range.
```
int
```

Print digits seprately

unsigned u = (unsigned) i;
if (i < 0) u = UINT_MAX - u;
printf("%u", u);

printf不打印unsigned int

1 个答案: