X处的堆块在Y处修改过去请求的大小为2 ERROR

Question

所以我在以下代码的末尾收到一个我完全不理解的错误：

    char* registerPointer = NULL;
    // parse the currentLine
    char** rt_rs_immediate;
    rt_rs_immediate = malloc(3 * sizeof(char*));
    if (rt_rs_immediate == NULL ) {
        fprintf(outputFilePointer, "no more memory");
        exit(1);
    }

    for (int i = 0; i < 3; i++) {
        rt_rs_immediate[i] = malloc(2 * sizeof(char));
        if (rt_rs_immediate[i] == NULL ) {
            fprintf(outputFilePointer, "no more memory");
            exit(1);
        }
    }

    int indexWithin_rt_rs_immediate = 0;
    registerPointer = strtok(currentLine, " $,\n\t");
    while (registerPointer != NULL ) {
        if (registerPointer == NULL || *registerPointer == '#') {
            break;
        } else {
            strcpy(rt_rs_immediate[indexWithin_rt_rs_immediate],
                    registerPointer);
            indexWithin_rt_rs_immediate++;
            registerPointer = strtok(NULL, " $,\n\t");
        }
    }
    free(registerPointer);

    // write to outputFile
    int immediate = atoi(rt_rs_immediate[2]);
    writeOutI_TypeInstruction(I_TypeInstruction, rt_rs_immediate[1],
            rt_rs_immediate[0], immediate, outputFilePointer);
    // free pointers created with malloc
    for (int i = 0; i < 3; i++) {
        free(rt_rs_immediate[i]); //<====================ERROR HERE!!!!!
    }
    free(rt_rs_immediate);

Answer 1

当while为regsiterPointer时，您的NULL循环终止。您可能想要检查该循环，因为您只有三个指针分配给rt_rs_immediate。如果你试图超过该循环中的三个指针，那可能会导致错误。