gdb提示：No symbol "expression" in current context.

void parseExpression(char* expression, char **expressionObjects){
        int expressionLength = strlen(expression);
        char operandString[20];
        int place = -1;
        int index = 0;
        int i;
        for(i = 0; i<expressionLength; i++){
                if(expression[i] == '+'||expression[i] == '-' || expression[i] == '*' || expression[i] == '/'){
                        strcpy(expressionObjects[i], operandString);//this line
                        index++;
                        expressionObjects[index][0] = expression[i];
                        index++;
                        place = -1;
                }
                else{
                        place++;
                        operandString[place] = expression[i];
                }
        }
        strcpy(expressionObjects[index+1], "end");
}

上面只是代码片段，在this line这个地方程序挂掉了，我在gdb中尝试打印i，expression，expressionObject[i],operandString，都提示No symbol "expression" in current context.
这是什么情况？ gdb c
[解决办法]
没看到expression有什么问题呢
但是你的strcpy函数拷贝啥？operandString内容是什么？你又没赋值又没初始化！
还有看看 拷贝会不会越界，每个够不够20个字节