Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$?

$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$
Given that \(a\), \(b\), and \(c\) are nonzero real numbers and \(a + b + c = 0\), we need to determine the possible values for the expression \(\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}\).

1. **Sign Analysis**:
   - The term \(\frac{a}{|a|}\) is the sign of \(a\), denoted as \(\operatorname{sgn}(a)\).
   - Similarly, \(\frac{b}{|b|} = \operatorname{sgn}(b)\) and \(\frac{c}{|c|} = \operatorname{sgn}(c)\).
   - The term \(\frac{abc}{|abc|}\) is the sign of the product \(abc\), denoted as \(\operatorname{sgn}(abc)\).

2. **Possible Sign Combinations**:
   - Since \(a + b + c = 0\), the sum of the signs of \(a\), \(b\), and \(c\) must be zero. This implies that either two of the numbers are positive and one is negative, or two are negative and one is positive.

3. **Case Analysis**:
   - **Case 1**: Two positive and one negative.
     - Without loss of generality, assume \(a > 0\), \(b > 0\), and \(c < 0\).
     - Then, \(\operatorname{sgn}(a) = 1\), \(\operatorname{sgn}(b) = 1\), and \(\operatorname{sgn}(c) = -1\).
     - The product \(abc\) will be negative since it involves two positive numbers and one negative number: \(\operatorname{sgn}(abc) = -1\).
     - Therefore, the expression becomes:
       \[
       \operatorname{sgn}(a) + \operatorname{sgn}(b) + \operatorname{sgn}(c) + \operatorname{sgn}(abc) = 1 + 1 - 1 - 1 = 0
       \]

   - **Case 2**: Two negative and one positive.
     - Without loss of generality, assume \(a < 0\), \(b < 0\), and \(c > 0\).
     - Then, \(\operatorname{sgn}(a) = -1\), \(\operatorname{sgn}(b) = -1\), and \(\operatorname{sgn}(c) = 1\).
     - The product \(abc\) will be negative since it involves two negative numbers and one positive number: \(\operatorname{sgn}(abc) = -1\).
     - Therefore, the expression becomes:
       \[
       \operatorname{sgn}(a) + \operatorname{sgn}(b) + \operatorname{sgn}(c) + \operatorname{sgn}(abc) = -1 - 1 + 1 - 1 = 0
       \]

4. **Conclusion**:
   - In both cases, the value of the expression \(\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}\) is \(0\).

The final answer is \(\boxed{0}\).